Page 12.2 Ex – 12.1
Q1.
Answer :
We have:
P(n): n(n + 1) is even.
Now,
P(3) = 3(3 + 1) = 12 (Even)
Therefore, P(3) is even.
Page 12.3 Ex – 12.1
Q2.
Answer :
We have:
P(n): n3+n is divisible by 3.Thus, we have:P(3) =33+3 =27+3=30 It is divisible by 3.Hence, P(3) is true.Now,P(4) =43+4=64+4=68 It is not divisible by 3.Hence, P(4) is not true.
Q3.
Answer :
We have:
P(n): 2n≥3nAlso,P(r) is true.∴2r≥3rTo Prove: P(r+1) is true.We have:2r≥3r⇒2r×2≥3r×2 Multiplying both sides by 2⇒2r+1≥6r∴2r+1≥3r+3 6r≥3r+3 for every r∈N.Hence, P(r+1) is true.
Q4.
Answer :
P(n): n2+n is even.Also, P(r) is true.Thus, r2+r is even.To prove: P(r+1) is true.Now,P(r+1)=(r+1)2+r+1 =r2+1+2r+r+1 =r2+3r+2 =r2+r+2r+2 =P(r)+2(r+1)P(r) is even.Also, 2(r+1) is even, as it is a multiple of 2.Therefore, P(r+1) is even and true.
Q5.
Answer :
Proved:
P(n)=n2+n is even for P(n) and P(n+1).Therefore, n2+n 2 is also even for all n∈N. Dividing an even number by 2 gives an even number.Thus, we have:P(n) =1+2+…+n =n(n+1)2 (Even for all n∈N))
Q6.
Answer :
P(n): n2-n+41 is prime.Now,P(1) =12-1+41=41 (prime)P(2)=22-2+41 =4-2+41 =43 (prime)P(3) =32-3+41=9-3+41=47 (prime)P(41)=412-41+41=1681 (not prime)Thus, we can say that P(1), P(2) and P(3) are true, but P(41) is not true.
Page 12.28 Ex – 12.2
Q1.
Answer :
Let P(n) be the given statement.
Now,
P(n) = 1 + 2 + 3 +…+ n =n(n+1)2
Step 1: P(1) =1 =1(1+1)2=1Hence, P(1) is true.Step 2: Let P(m) be true.Then,1+2+3+…+m=m(m+1)2We shall now prove that P(m+1) is true.i.e., 1+2+…+(m+1) =(m+1)(m+2)2Now, 1+2+…+m =m(m+1)2⇒1+2+…m+m+1 =m(m+1)2+m+1 Adding m+1 to both sides =m2+m+2m+22 =(m+1)(m+2)2Hence, P(m+1) is true.
By the principle of mathematical induction, P(n) is true for all n∈N.
Q2.
Answer :
Let P(n) be the given statement.
Now,
P(n) =12+22+32+…+n2=n(n+1)(2n+1)6Step 1:P(1) =12=1(1+1)(2+1)6=66=1Hence, P(1) is true.Step 2:Let P(m) be true. Then,12+22+…+m2=m(m+1)(2m+1)6We shall now prove that P(m+1) is true.i.e., 12+22+32+…+(m+1)2=(m+1)(m+2)(2m+3)6Now,P(m) = 12+22+32+…+m2=m(m+1)(2m+1)6⇒12+22+32+…+m2+(m+1)2=m(m+1)(2m+1)6+(m+1)2 Adding (m+1)2 to both sides⇒12+22+32+…+(m+1)2=m(m+1)(2m+1)+6(m+1)26=(m+1)(2m2+m+6m+6)6=(m+1)(m+2)(2m+3)6Hence, P(m+1) is true.By the principle of mathematical induction, the given statement is true for all n∈N
Q3.
Answer :
Let P(n) be the given statement.
Now,
P(n) =1+3+32+…+3n-1=3n-12Step 1:P(1) =1 =31-12=22=1Hence, P(1) is true.Step 2:Let P(m) is true.Then,1+3+32+…+3m-1=3m-12We shall prove that P(m+1) is true.That is, 1+3+32+…+3m=3m+1-12Now, we have:1+3+32+…+3m-1=3m-12⇒1+3+32+…+3m-1+3m=3m-12+3m Adding 3m to both sides⇒1+3+32+…+3m=3m-1+2×3m2=3m(1+2)-12=3m+1-12Hence, P(m+1) is true.By the principle of mathematical induction, P(n) is true for all n∈N.
Q4.
Answer :
Let P(n) be the given statement.
Now,
P(n) =11.2+12.3+13.4+…+1n(n+1)=nn+1Step 1:P(1) =11.2=12=11+1Hence, P(1) is true.Step 2:Let P(m) be true.Then,11.2+12.3+13.4+…+1m(m+1)=mm+1We shall now prove that P(m+1) is true.i.e., 11.2+12.3+13.4+…+1(m+1)(m+2)=m+1m+2Now,P(m)= 11.2+12.3+13.4+…+1m(m+1)=mm+1⇒11.2+12.3+13.4+…+1m(m+1)+1(m+1)(m+2)=mm+1+1(m+1)(m+2) Adding 1(m+1)(m+2) to both sides⇒11.2+12.3+13.4+…+1(m+1)(m+2)=m2+2m+1(m+1)(m+2)=(m+1)2(m+1)(m+2)=m+1m+2Therefore, P(m+1) is true.By the principle of mathematical induction, P(n) is true for all n∈N.
Q5.
Answer :
Let P(n) be the given statement.
Now,
P(n) = 1+3+5+…+(2n-1)=n2Step 1:P(1) =1 = 12Hence, P(1) is true.Step 2 :Let P(m) be true.Then,1+3+5+…+(2m-1) = m2To prove: P(m+1) is true.i.e., 1+3 +5+…+2m+1-1=m+12 ⇒1+3 +5+…+2m+1=m+12Now, we have:1+3+5+…+(2m-1) =m2⇒1+3+…+(2m-1)+(2m+1) =m2+2m+1 Adding 2m+1 to both sides⇒1+3+5+…+(2m+1) =(m+1)2Hence, P(m+1) is true.By the principle of mathematical induction, P(n) is true for all n∈N.
Q6.
Answer :
Let P(n) be the given statement.
Now,
P(n) =12.5+15.8+18.11+…+1(3n-1)(3n+2)=n6n+4Step 1:P(1) =12.5=110=16+4Hence, P(1) is true.Step 2:Let P(m) be true. Then,12.5+15.8+18.11+…+1(3m-1)(3m+2)=m6m+4To prove: P(m+1) is true.i.e.,12.5+15.8+…+1(3m+2)(3m+5)=m+16m+10Thus, we have: 12.5+15.8+18.11+…+1(3m-1)(3m+2)=m6m+4⇒12.5+15.8+…+1(3m-1)(3m+2)+1(3m+2)(3m+5)=m6m+4+1(3m+2)(3m+5) Adding 1(3m+2)(3m+5) to both sides⇒12.5+15.8+…+1(3m+2)(3m+5)=3m2+5m+22(3m+2)(3m+5)=(3m+2)(m+1)2(3m+2)(3m+5)=m+16m+10Thus, P(m+1) is true.By the principle of mathematical induction, P(n) is true for all n∈N.
Q7.
Answer :
Let P(n) be the given statement.
Now,
P(n) =11.4+14.7+17.10+…+1(3n-2)(3n+1)=n3n+1Step 1:P(1) =11.4=14=13×1+1Hence, P(1) is true.Step 2:Let P(m) be true.i.e.,11.4+14.7+…+1(3m-2)(3m+1)=m3m+1To prove: P(m+1) is true.i.e.,11.4+14.7+…+1(3m+1)(3m+4)=m+13m+4Now,P(m)=11.4+14.7+…+1(3m-2)(3m+1)=m3m+1⇒11.4+14.7+…+1(3m-2)(3m+1)+1(3m+1)(3m+4)=m3m+1+1(3m+1)(3m+4) Adding 1(3m+1)(3m+4) to both sides⇒11.4+14.7+…+1(3m+1)(3m+4)=3m2+4m+1(3m+1)(3m+4)=(3m+1)(m+1)(3m+1)(3m+4)=m+13m+4Thus, P(m+1) is true.By the principle of mathematical induction, P(n) is true for all n∈N.
Q8.
Answer :
Let P(n) be the given statement.
Now,
P(n)=13.5+15.7+17.9+…+1(2n+1)(2n+3)=n3(2n+3)Step 1: P(1)=13.5=115=13(2+3)Hence, P(1) is true.Step 2:Let P(m) be true. Then,13.5+15.7+17.9+…+1(2m+1)(2m+3)=m3(2m+3)To prove: P(m+1) is true.That is,13.5+15.7+17.9+…+1(2m+3)(2m+5)=m+13(2m+5)Now, P(m) = 13.5+15.7+17.9+…+1(2m+1)(2m+3)=m3(2m+3)⇒13.5+15.7+…+1(2m+1)(2m+3)+1(2m+3)(2m+5)=m3(2m+3)+1(2m+3)(2m+5) Adding 1(2m+3)(2m+5) to both sides⇒13.5+15.7+…+1(2m+3)(2m+5)=2m2+5m+33(2m+3)(2m+5)=(2m+3)(m+1)3(2m+3)(2m+5)=m+132m+5Thus, P(m+1) is true.By the principle of mathematical induction, P(n) is true for all n∈N.
Q9.
Answer :
Let P(n) be the given statement.
Now,
P(n) =13.7+17.11+111.15+…+1(4n-1)(4n+3)=n3(4n+3)Step 1:P(1) =13.7=121=13(4+3)Hence, P(1) is true.Step 2:Let P(m) is true. Then,13.7+17.11+…+1(4m-1)(4m+3)=m3(4m+3)To prove: P(m+1) is true.That is,13.7+17.11+…+1(4m+3)(4m+7)=m+13(4m+7)Now,P(m) =13.7+17.11+…+1(4m-1)(4m+3)=m3(4m+3)⇒13.7+17.11+…+1(4m-1)(4m+3)+1(4m+3)(4m+7)=m3(4m+3)+1(4m+3)(4m+7) Adding 1(4m+3)(4m+7) to both sides⇒13.7+17.11+…+1(4m+3)(4m+7)=4m2+7m+33(4m+3)(4m+7)=(4m+3)(m+1)3(4m+3)(4m+7)=m+13(4m+7)Thus, P(m+1) is true.By the principle of mathematical induction, P(n) is true for all n∈N.
Q10.
Answer :
Let P(n) be the given statement.
Now,
P(n) =1.2+2.22+3.23+…+n.2n=(n-1)2n+1+2Step 1:P(1)=1.2=2=(1-1)21+1+2Thus, P(1) is true.Step 2:Let P(m) be true.Then,1.2+2.22+…+m.2m=(m-1)2m+1+2To prove: P(m+1) is true.That is,1.2+2.22+…+(m+1)2m+1=m.2m+2+2Now, P(m) =1.2+2.22+…+m.2m=(m-1)2m+1+2⇒1.2+2.22+…+m.2m+(m+1).2m+1=(m-1)2m+1+2+(m+1).2m+1 Adding (m+1).2m+1 to both sides⇒P(m+1)=2m.2m+1+2=m.2m+2+2Thus, P(m+1) is true.By the principle of mathematical induction, P(n) is true for all n∈N.
Q11.
Answer :
Let P(n) be the given statement.
Now,
P(n) =2+5+8+…+(3n-1)=12n(3n+1)Step 1:P(1)=2=12×1(3+1) Hence, P(1) is true.Step 2:Let P(m) be true.Then,2+5+8+…+(3m-1)=12m(3m+1)To prove: P(m+1) is true.That is,2+5+8+…+(3m+2)=12(m+1)(3m+4)P(m) is equal to:2+5+8+…+(3m-1)=12m(3m+1)Thus, we have:2+5+8+…+(3m-1)+(3m+2)=12m(3m+1)+(3m+2) Adding (3m+2) to both sides⇒2+5+8+…+(3m+2)=12(3m2+m+6m+4)=12(3m2+7m+4)⇒2+5+8+…+(3m+2)=12(3m+4)(m+1)Thus, P(m+1) is true.By the principle of mathematical induction, P(n) is true for all n∈N.
Q12.
Answer :
Let P(n) be the given statement.
Now,
P(n) = 1.3+2.4+3.5+…+n.(n+2)=16n(n+1)(2n+7)Step 1:P(1)= 1.3 =3 = 16×1(1+1)(2×1+7)Hence, P(1) is true.Step 2:Let P(m) be true.Then,1.3+2.4+…+m.(m+2)=16m(m+1)(2m+7)To prove: P(m+1) is true.That is,1.3+2.4+…+(m+1)(m+3)=16(m+1)(m+2)(2m+9)P(m) is equal to 1.3+2.4+…+m(m+2)=16m(m+1)(2m+7).Thus, we have:1.3+2.4+…+m(m+2)+(m+1)(m+3) = 16m(m+1)(2m+7)+(m+1)(m+3) Adding (m+1)(m+3) to both sides⇒1.3+2.4+…+(m+1)(m+3)=16(m+1)2m2+7m+6m+18 =16(m+1)(2m2+13m+18) =16(m+1)(2m+9)(m+2)Thus, P(m+1) is true.By the principle of mathematical induction, P(n) is true for all n∈N.
Q13.
Answer :
Let P(n) be the given statement.
Now,
P(n)=1.3+3.5+5.7+…+(2n-1)(2n+1)=n(4n2+6n-1)3Step 1:P(1)=1.3 =3 =1(4×12+6×1-1)3Hence, P(1) is true.Step 2:Let P(m) be true.Then,1.3+3.5+…+(2m-1)(2m+1)=m(4m2+6m-1)3To prove: P(m+1) is true.That is,1.3+3.5+…+(2m+1)(2m+3)=(m+1)4(m+1)2+6m+1-13Now, P(m) is equal to: 1.3+3.5+…+(2m-1)(2m+1)=m(4m2+6m-1)3⇒1.3+3.5+…+(2m-1)(2m+1)+(2m+1)(2m+3)=m(4m2+6m-1)3+(2m+1)(2m+3) Adding (2m+1)(2m+3) to both sides⇒P(m+1)=m(4m2+6m-1)+3(4m2+8m+3)3⇒P(m+1)=4m3+6m2-m+12m2+24m+93=4m3+18m2+23m+93⇒P(m+1)=4m(m2+2m+1)+10m2+19m+93 =4m(m+1)2+(10m+9)(m+1)3 =(m+1)4m(m+1)+10m+93 =(m+1)3(4m2+8m+4+6m+5) =(m+1)4(m+1)2+6m+1-13Thus, P(m+1) is true.By the principle of mathematical induction, P(n) is true for all n∈N.
Q14.
Answer :
Let P(n) be the given statement.
Now,
P(n)=1.2+2.3+3.4+…+n(n+1)=n(n+1)(n+2)3Step 1:P(1)=1.2=2=1(1+1)(1+2)3Hence, P(1) is true.Step 2:Let P(m) be true.Then,1.2+2.3+…+m(m+1)=m(m+1)(m+2)3To prove: P(m+1) is true.That is,1.2+2.3+…+(m+1)(m+2)=(m+1)(m+2)(m+3)3Now, P(m) is 1.2+2.3+…+m(m+1)=m(m+1)(m+2)3⇒1.2+2.3+…+m(m+1)+(m+1)(m+2)=m(m+1)(m+2)3+(m+1)(m+2)⇒P(m+1)=(m+1)(m+2)(m+3)3Thus, P(m+1) is true.By the principle of mathematical induction, P(n) is true for all n∈N.’
Q15.
Answer :
Let P(n) be the given statement.
Now,
P(n): 12+14+18+…+12n=1-12nStep 1: P(1)=12=1-121Thus, P(1) is true.Step 2:Suppose P(m) is true.Then,12+14+…+12m=1-12mTo show: P(m+1) is true whenever P(m) is true.That is,12+14+…+12m+1=1-12m+1Now, P(m) is true.Thus, we have:12+14+…+12m=1-12m⇒12+14+…+12m+12m+1=1-12m+12m+1 Adding 12m+1 to both sides⇒P(m+1)=1-12m+12m.2=1-12 m1-12=1-12m+1Thus, P(m+1) is true.
By the principle of mathematical induction, P(n) is true for all n∈N.
Q16.
Answer :
Let P(n) be the given statement.
Now,
P(n)=12+32+52+…+(2n-1)2=13n(4n2-1)Step 1: P(1)=12=1=13×1×(4-1)Hence, P(1) is true.Step 2:Let P(m) be true.Then,12+32+…+(2m-1)2=13m(4m2-1)To prove: P(m+1) is true whenever P(m) is true.That is, 12+32=…+(2m+1)2=13(m+1)4(m+1)2-1We know that P(m) is true.Thus, we have:12+32+…+(2m-1)2=13m(4m2-1)⇒12+32+…+(2m-1)2+(2m+1)2=13m(4m2-1)+(2m+1)2 Adding (2m+1)2 to both sides⇒P(m+1)=134m3-m+12m2+12m+3⇒P(m+1)=13(4m3-m+8m2+4m+4m2+8m+3) =13(m+1)(4m2+8m+3) =13(m+1)(4(m+1)2-1)Thus, P(m+1) is true.By the principle of mathematical induction, P(n) is true for all n∈N.
Q17.
Answer :
Let P(n) be the given statement.
Now,
P(n)= a+ar+ar2+…+arn-1=arn-1r-1, r≠1Step 1:P(1)=a=ar1-1r-1Hence, P(1) is true.Step 2:Suppose P(m) is true. Then,a+ar+ar2+…+arm-1=arm-1r-1, r≠1To show: P(m+1) is true whenever P(m) is true.That is,a+ar+ar2+…+arm=arm+1-1r-1, r≠1We know that P(m) is true.Thus, we have:a+ar+ar2+…+arm-1=arm-1r-1⇒a+ar+ar2+…+arm-1+arm=arm-1r-1+arm Adding arm to both sides⇒P(m+1)=arm-1+r.rm-rmr-1⇒P(m+1)=arm+1-1r-1, r≠1Thus, P(m+1) is true.By the principle of mathematical induction, P(n) is true for all n∈N.
Q18.
Answer :
Let P(n) be the given statement.
Now,
P(n): a+(a+d)+(a+2d)+…+(a+(n-1)d)=n22a+(n-1)dStep1:P(1)= a= 12(2a+(1-1)d)Hence, P(1) is true.Step 2:Suppose P(m) is true.Then,a+(a+d)+…+(a+(m-1)d)=m22a+(m-1)dWe have to show that P(m+1) is true whenever P(m) is true.That is,a+(a+d)+…+(a+md)=(m+1)22a+mdWe know that P(m) is true.Thus, we have:a+(a+d)+…+(a+(m-1)d)=m22a+(m-1)d⇒a+(a+d)+…+(a+(m-1)d)+(a+md)=m22a+(m-1)d+(a+md) Adding (a+md) to both sides⇒P(m+1)=122am+m2d-md+2a+2md⇒P(m+1)=122a(m+1)+md(m+1) =12(m+1)(2a+md)Thus, P(m+1) is true.By the principle of mathematical induction, P(n) is true for all n∈N.
Q19.
Answer :
Let P(n) be the given statement.
Now,
P(n): 7+77+777+…+777…n digits…7=781(10n+1-9n-10)Step(1): P(1) = 7 =781(102-9-10)=781×81 Thus, P(1) is true.Step 2: Let P(m) be true.Then, 7+77+777+…+777…m digits…7=781(10m+1-9m-10)We need to show that P(m+1) is true whenever P(m) is true.
Now, P(m + 1) = 7 + 77 + 777 +….+ 777…(m + 1) digits…7
This is a geometric progression with n= m+1.∴Sum P(m+1): =799+99+999+…m+1term=7910-1+100-1+…(m+1) term=7910+100+1000+…(m+1) term -(1+1+1…m+1 times…+1=791010m+1-19-m+1=78110m+2-9m-19Thus, P(m+1) is true.By the principle of mathematical induction, Pn is true for all n∈N.
Page 12.29 Ex – 12.2
Q20.
Answer :
Let P(n) be the given statement.
Now,
P(n): n77+n55+n33+n22-37210n is a positive integer.Step 1: P(1)=17+15+13+12-37210=30+42+70+105-37210=210210=1 It is a positive integer.Thus, P(1) is true.Step 2:Let P(m) be true.Then, m77+m55+m33+m22-37210m is a positive integer.Let m77+m55+m33+m22-37210m=λ for some λ∈positive N.To show: Pm+1 is a positive integer.Now,P(m+1)=m+177+m+155+m+133+m+122-37210m+1=17m7+7m6+21m5+35m4+35m3+21m2+7m+1+15m5+5m4+10m3+10m2+5m+1+13m3+3m2+3m+1+12m2+2m+1-37210m-37210 =m77+m55+m33+m22-37210m +m6+3m5+6m4+7m3+6m2+4m=λ+ m6+3m5+6m4+7m3+6m2+4mIt is a positive integer, as λ is a positive integer.Thus, Pm+1 is true,By the principle of mathematical induction, P(n) is true for all n∈N.
Q21.
Answer :
Let P(n) be the given statement.
Now,
P(n): n1111+n55+n33+62165n is a positive integer for all n∈N.Step 1:P(1) =111+15+13+62165=15+33+55+62165=165165=1 It is certainly a positive integer.Hence, P(1) is true.Step2:Let P(m) be true.Then, m1111+m55+m33+62165m is a positive integer.Now, let m1111+m55+m33+62165m =λ, where λ∈N is a positive integer.We have to show that P(m+1) is true whenever P(m) is true.To prove: (m+1)1111+(m+1)55+(m+1)33+62165(m+1) is a positive integer.Now,(m+1)1111+(m+1)55+(m+1)33+62165(m+1)=111m11+11m10+55m9+165m8+330m7+462m6+462m5+330m4+165m3+55m2+11m+1+15m5+5m4+10m3+10m2+5m+1+13m3+3m2+3m+1+62165m+62165=m1111+m55+m33+62165m+m10+5m9+15m8+30m7+42m6+42m5+31m4+17m3+8m2+3m+111+15+13+6105=λ+m10+5m9+15m8+30m7+42m6+42m5+31m4+17m3+8m2+3m+1It is a positive integer.Thus, P(m+1) is true.By the principle of mathematical induction, P(n) is true for all n∈N.
Q22.
Answer :
Let P(n) be the given statement.
Now,
P(n): 52n-1 is divisible by 24 for all n∈N.Step 1: P(1)=52-1=25-1=24 It is divisible by 24.Thus, P(1) is true.Step 2:Let P(m) be true.Then, 52m-1 is divisible by 24.Now, let 52m-1 = 24λ, where λ∈N.We need to show that P(m+1) is true whenever P(m) is true.Now,P(m+1) =52m+2-1 =52m52-1 =25(24λ+1) -1 =600λ+24 =24(25λ+1)It is divisible by 24.Thus, P(m+1) is true.By the principle of mathematical induction, P(n) is true for all n∈N.
Q23.
Answer :
Let P(n) be the given statement.
Now,
P(n): 32n+7 is divisible by 8 for all n∈N.Step 1:P(1)= 32+7=9+7=16 It is divisible by 8.Step 2: Let P(m) be true.Then, 32m+7 is divisible by 8.Thus, 32m+7=8λ for some λ∈N. …(1)We need to show that P(m+1) is true whenever P(m) is true.Now,P(m+1) =32m+2+7 =32m.9 +7 =(8λ-7).9+7 From (1) =72λ-63+7 =72λ-56 =8(9λ-7) It is a multiple of 8.Thus, P(m+1) is divisible by 8.By the principle of mathematical induction, P(n) is true for all n∈N.
Q24.
Answer :
Let P(n) be the given statement.
Now,
P(n): 52n+2-24n-25 is divisible by 576 for all n∈N.Step 1:P(1)= 52+2-24-25=625-49=576 It is divisible by 576.Thus, P(1) is true.Step2: Let P(m) be true.Then,52m+2-24m-25 is divisible by 576.Let 52m+2-24m-25 =576λ, where λ∈N.We need to show that P(m+1) is true whenever P(m) is true.Now, P(m+1)=52m+4-24(m+1)-25=52×(576λ+24m+25)-24m-49=25×576λ+600m+625-24m-49=25×576λ+576m+576=576(25λ+m+1) It is divisible by 576.Thus, P(m+1) is true.By the principle of mathematical induction, P(n) is true for all n∈N.
Q25.
Answer :
Let P(n) be the given statement.
Now,
P(n): 32n+2-8n-9 is divisible by 8 for all n∈N.Step 1:P(1)=32+2-8-9=81-17=64 It is divisible by 8.Thus, P(1) is true.Step(2):Let P(m) be true.Then, 32m+2-8m-9 is divisible by 8.Let:32m+2-8m-9 =8λ where λ∈N …(1)We need to show that P(m+1) is true whenever P(m) is true.Now,P(m+1)=32m+4-8m+1-17=(8λ+8m+9) -8m -8-17 From (1)=8λ -16 =8(λ-1) It is divisible by 8.Thus, P(m+1) is true.By the principle of mathematical induction, P(n) is true for all n∈N.
Q26.
Answer :
Let P(n) be the given statement.
Now,
P(n): (ab)n=anbn for all n∈N.Step 1:P(1):(ab)1=a1b1=abThus, P(1) is true.Step 2:Let P(m) be true.Then,(ab)m=ambmWe need to show that P(m+1) is true whenever P(m) is true.Now, P(m+1): (ab)m+1=(ab)m. ab =ambm.ab =ama.bmb =am+1bm+1Hence, P(m+1) is true.By the principle of mathematical induction, P(n) is true for all n∈N.
Q27.
Answer :
Let P(n) be the given statement.
Now,
P(n):n(n+1)(n+5) is a multiple of 3.Step1:P(1): 1(1+1)(1+5)=12 It is a multiple of 3.Hence, P(1) is true.Step2: Let Pm be true.Then, mm+1m+5 is a multiple of 3.Suppose mm+1m+5=3λ, where λ∈N.We have to show that Pm+1 is true whenever P(m) is true.Now,P(m+1)=m+1m+2m+6 =mm+1m+6+2m+1m+6 =mm+1m+5+1+2m+1m+6 =mm+1m+5+m(m+1)+2m+1m+6 =3λ+m+1m+2m+6 From P(m) =3λ+3m+1m+2It is clearly a multiple of 3.Thus, P(m+1) is true.By the principle of mathematical induction, P(n) is true for all n∈N.
Q28.
Answer :
Let P(n) be the given statement.
Now,
P(n): 72n+23n-3.3n-1 is divisible by 25.Step1: P(1): 72+23-3.31-1=49+1=50 It is divisible by 25.Thus, P(1) is trueStep2: Let Pm be true.Now,72m+23m-3.3m-1 is divisible by 25.Suppose: 72m+23m-3.3m-1= 25λ …(1)We have to show that Pm+1 is true whenever P(m) is true.Now, Pm+1=72m+2+23m.3m =72m+2+72.23m-3.3m-1-72.23m-3.3m-1+23m.3m =7272m+23m-3.3m-1+23m.3m1-4924 =72×25λ-23m.3m×2523.31 Using (1) =2549λ-23m-3.3m-1 It is divisible by 25.Thus, Pm+1 is true.By the principle of mathematical induction, P(n) is true for all n∈N.
Q29.
Answer :
Let P(n) be the given statement.
Now,
P(n): 2.7n+3.5n-5 is divisible by 24.Step1:P(1): 2.71+3.51-5 = 24 It is divisible by 24.Thus, P1 is true.Step2: Let Pm be true.Then,2.7m+3.5m-5 is divisible by 24.Suppose:2.7m+3.5m-5=24λ …1We need to show that Pm+1 is true whenever Pm is true.Now, Pm+1=2.7m+1+3.5m+1-5 =2.7m+1+(24λ+5-2.7m)5-5 =2.7m+1+120λ+25-10.7m-5 =2.7m.7-10.7m+120λ+24-4 =7m 14-10+120λ+24-4 =7m.4+120λ+24-4 =47m-1+245λ+1 =4×6μ+24(5λ+1) Since 7m-1 is a multiple of 6 for all n∈N, 7m-1=μ. =24(μ+5λ+1) It is a multiple of 24.Thus, P(m+1) is true.By the principle of mathematical induction, P(n) is true for n∈N.
Q30.
Answer :
Let P(n) be the given statement.
Now,
Pn: 11n+2+122n+1 is divisible by 133.Step1: P1=111+2+122+1=1331+1728=3059 It is divisible by 133.Step2:Let Pm be divisible by 133.Now,11m+2+122m+1 is divisible by 133.Suppose:11m+2+122m+1=133λ …(1)We shall show that Pm+1 is true whenever Pm is true.Now, Pm+1=11m+3+122m+3 =11m+2.11+122m+1.122+11.122m+1-11.122m+1 =1111m+2+122m+1+122m+1144-11 =11.133λ+122m+1.133 From (1) =13311λ+122m+1 It is divisible by 133.Thus, Pm+1 is true.By the principle of mathematical induction, P(n) is true for all n∈N.
Q31.
Answer :
We need to prove 12tanx2+14tanx4+…+12ntanx2n=12ncotx2n-cot x for all n ∈ N and 0<x<π2 using mathematical induction.
For n = 1,
LHS = 12tanx2
and
RHS=12cotx2-cotx=12tanx2-1tanx⇒RHS=12tanx2-12tanx21-tan2x2⇒RHS=12tanx2-1-tan2x22 tanx2=1-1+tan2x22tanx2=tanx22
Therefore, the given relation is true for n = 1.
Now, let the given relation be true for n = k.
We need to prove that the given relation is true for n = k + 1.
∴12tanx2+14tanx4+…+12ktanx2k=12kcotx2k-cot x
Now,
12tanx2+14tanx4+…+12ktanx2k+12k+1tanx2k+1=12kcotx2k-cot x+12k+1tanx2k+1
Let:
L=12kcotx2k-cot x+12k+1tanx2k+1.
⇒L=12kcotx2k+12k+1tanx2k+1-cot x⇒L=12ktanx2k+12k+1tanx2k+1-cot x⇒L=12ktan2x2k+1+12k+1tanx2k+1-cot x⇒L=12k×2tanx2k+11-tan2x2k+1+12k+1tanx2k+1-cot x⇒L=1-tan2x2k+12k+1tanx2k+1+12k+1tanx2k+1-cot x
⇒L=1-tan2x2k+1+tan2x2k+12k+1tanx2k+1-cot x=12k+1cotx2k+1-cot x
Now,
12tanx2+14tanx4+…+12ktanx2k+12k+1tanx2k+1=12k+1cotx2k+1-cot x
Thus, 12tanx2+14tanx4+…+12ntanx2n=12ncotx2n-cot x for all n ∈ N and 0<x<π2.
Q32.
Answer :
Pn:2n≥3nWe know that Pr is true.Thus, we have:2r≥3rTo show: P(r+1) is true.We know:P(r) is true.∴2r≥3r⇒2r.2≥3r.2 Multiplying both sides by 2⇒2r+1≥6r⇒2r+1≥3r+3r=2r+1≥3r+3 Since 3r≥3 for all r∈N=2r+1≥3r+1 Hence, P(r+1) is true.However, we cannot conclude that Pn is true for all n∈N.P(1): 21≱3.1Therefore, Pn is not true for all n∈N.
Q33.
Answer :
Let P(n) be the given statement.
Thus, we have:
Pn:2n!22nn!2≤13n+1Step1: P(1): 2!22.1=12≤13+1Thus, P(1) is true.Step2: Let P(m) be true. Thus, we have:2m!22mm!2≤13m+1We need to prove that P(m+1) is true.
Now,
P(m+1): (2m+2)!22m+2(m+1)!2=2m+22m+12m!22m.22m+12m!2⇒(2m+2)!22m+2(m+1)!2≤2m!22mm!2×2m+22m+122m+12⇒(2m+2)!22m+2(m+1)!2≤2m+12m+13m+1
⇒2m+2!22m+2m+1!2≤2m+124m+123m+1⇒2m+2!22m+2m+1!2≤4m2+4m+1×3m+443m3+7m2+5m+13m+4⇒2m+2!22m+2m+1!2≤12m3+28m2+19m+412m3+28m2+20m+43m+4∵12m3+28m2+19m+412m3+28m2+20m+4<1∴2m+2!22m+2m+1!2<13m+4
Thus, P(m + 1) is true.
Hence, by mathematical induction (2n)!22n(n!)2≤13n+1 is true for all n ∈ N
Q34.
Answer :
Let P(n) be the given statement.
Thus, we have:
Pn:1+14+19+…+1n2<2-1nStep1: P(2):122=14<2-12Thus, P2 is true. We have not taken n =1 because it is not possible. We will start this function from n=2 onwards.Step2: Let Pm be true. Now,1+14+19+…+1m2<2-1mWe need to prove that P(m+1) is true.We know that P(m) is true.Thus, we have:1+14+19+…+1m2<2-1m⇒1+14+19+…+1m2+1m+12<2-1m+1m+12 Adding 1(m+1)2 to both sides⇒Pm+1<2-1m+1 ∵ m+12>m+1, 1m+12<1m+1⇒1m-1m+12<1m+1 as m<m+1 Thus, Pm+1 is true.By principle of mathematical induction, P(n) is true for all n∈N, n≥2.
Q35.
Answer :
Let P(n) be the given statement.
Now,
P(n): x2n-1+y2n-1 is divisible by x+y.Step1:P(1):x2-1+y2-1=x+y is divisible by x+yStep2:Let P(m) be true.Also,x2m-1+y2m-1 is divisible by x+y.Suppose: x2m-1+y2m-1=λx+y where λ∈N …(1)We shall show that Pm+1 is true whenever Pm is true.Now, Pm+1=x2m+1+y2m+1 =x2m+1+y2m+1-x2m-1.y2+x2m-1.y2 =x2m-1×2-y2+y2x2m-1+y2m-1 From (1) =x2m-1×2-y2+y2.λx+y =x+yx2m-1x-y+λy2 [It is divisible by (x+y).]Thus, Pm+1 is true.By the principle of mathematical induction, P(n) is true for all n∈N.
Q36.
Answer :
Let:Pn:an-Aan+A=a1-Aa1+A2n-1Step I P(1):a1-Aa1+A=a1-Aa1+A21-1 (which is true)P(2):a2-Aa2+A=12a1+Aa1-A12a1+Aa1+A=a1+Aa1-2Aa1+Aa1+2A=a1+A-2Aa1+A+2A=a1-Aa1+A22-1Thus, P(1) and P(2) are true.Step II Let P(k) be true.Now, ak-Aak+A=a1-Aa1+A2k-1 …..(i)andP(k+1):ak+1-Aak+1+A=12ak+Aak-A12ak+Aak+A=ak+Aak-2Aak+Aak+2A=ak-Aak2ak+Aak2=ak-Aakak+Aak2=ak-Aak+A2=a1-Aa1+A2k-12=a1-Aa1+A2k=a1-Aa1+A2k+1-1 Thus, P(k+1) is also true.
Q37.
Answer :
Let P(n) be the given statement.
P(n): sin x+sin 3x+…+sin2n-1x=sin2nxsin xStep 1:P(1): sinx=sin2xsinx Thus, P(1) is true. Step 2:Let P(m) be true.∴ sin x+sin 3x+…+sin2m-1x=sin2mxsin xWe shall show that P(m+1) is true. We know that P(m) is true. ∴ sin x+sin 3x+…+sin (2m-1) =sin2mxsin x⇒sin x+sin 3x+…sin (2m-1)x+ sin (2m+1)x=sin2mxsin x+sin (2m+1)x Adding sin (2m+1)x to both the sides⇒P(m+1)x=sin2mx+sin xsin mxcosm+1x+sinm+1x cosmxsin x =sin2mx+sin xsin mxcos mxcos x-sin2mxsinx+sin mxcos xcos mx+cos2mxsin xsin x =sin2mx+2sin xcos xcos mx-sin2x sin2mx+cos2mx sin2xsin x =sin2mx1-sin2x+2sin xcos xcos mx+cos2mx sin2xsin x =sin2mxcos2x+2sin xcos xcos mx+cos2mx sin2xsin x =sin mx cos x+cos mx sinx2sin x =sinm+12sin xHence, P(m+1) is true. By the principle of mathematical induction, the given statement P(n) is true for all n∈N.
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