Free Principle of Mathematical Induction 02 Practice Test - 11th Grade - Commerce 

Putting n = 1 in ${11}^{n+2}+{12}^{2n+1}$

We get, ${11}^{1+2}+122\times 1+1$ = ${11}^3+{12}^3$ = 3059, which

is divisible by 133.

Putting n = 2 in $3^{2n}$ - 2n + 1 then,

$3^{2\times 2}$ - 2×2+1 = 81 - 4 + 1 = 78, which is divisible

by 2.

$P\left(n\right):{2.7}^n+{3.5}^n-5P\left(1\right):=14+15-5=24$

Suppose 24 divides $P\left(n\right):{2.7}^n+{3.5}^n-5$
P(1) is true.
Suppose P(k) is true.
${2.7}^k+{3.5}^k-5$ is divisible by 24.
$\therefore {2.7}^k+{3.5}^k-5=24m{2.7}^{k+1}+{3.5}^{k+1}-5=7.\left({2.7}^k\right)+5.\left({3.5}^k\right)-5=5({2.7}^k+{3.5}^k-5)+{\mathrm{2.2.7}}^k+20=5({2.7}^k+{3.5}^k-5)+{4.7}^k+20=5\left(24m\right)+{4.7}^k+20$

Now, prove that $P_1\left(n\right):{4.7}^n+20isdivisibleby24$
$P_1\left(1\right):28+20=48\to divisibleby24P_1\left(k\right):{4.7}^k+20isdivisibleby24\to assumetrue\Rightarrow {4.7}^k+20=24qNow,{4.7}^{k+1}+20=7.\left({4.7}^k\right)+20={\mathrm{6.4.7}}^k+({4.7}^k+20)={24.7}^k+24q{P}_1(k+1)istrue$

$\therefore {4.7}^n+20isdivisibleby24$

$\therefore 5\left(24m\right)+{4.7}^k+20$ is divisible by 24
P(k) is true $\Rightarrow$ P(k+1) is true
Hence, P(n) is true.

$P\left(n\right):{4.7}^n+20isdivisibleby24$
$P\left(1\right):28+20=48\to divisibleby24P\left(k\right):{4.7}^k+20isdivisibleby24\to assumetrue\Rightarrow {4.7}^k+20=24qNow,{4.7}^{k+1}+20=7.\left({4.7}^k\right)+20={\mathrm{6.4.7}}^k+({4.7}^k+20)={24.7}^k+24qP(k+1)istrue$

$\therefore P\left(n\right)istrue{4.7}^n+20isdivisibleby24$

$P\left(n\right):(n+1)(n+2)(n+3)P\left(1\right):2\times 3\times 4=24\to \text{divisible by 6}P\left(n\right):(n+1)(n+2)(n+3)\text{is divisible by 6}$

P(1) is true. Assume P(k) is true
$\Rightarrow (k+1)(k+2)(k+3)=6mP(k+1):(k+2)(k+3)(k+4)=(k+1)(k+2)(k+3)+3(k+2)(k+3)$

Since (k+2) and (k+3) are two consecutive natural numbers, one of them must be even i.e. their product is even, say 2q
$\therefore (k+2)(k+3)(k+4)=(k+1)(k+2)(k+3)+3.2q=6m+6q=6(m+q)\to \text{divisible by 6}\Rightarrow \text{P(k+1) is true}\Rightarrow \text{P(n) is true}$

Substituting n=1, we get
$a^1-b^1=a-b$
Let $P\left(n\right):a^n-b^n\text{is divisible by a-b}$
P(1) is true
Assume P(k) is true
$a^k-b^k\text{is divisible by a-b}{a}^k-b^k=m(a-b)Now,a^{k+1}-b^{k+1}=a.(a^k-b^k)+(a-b)b^k=(am+b^k)(a-b)\to \text{divisible by a-b}$
P(k+1) is true

Hence, P(n) is true.

Substitute n=1 in $2^{3n}-7n-1$, we get
$2^3-7-1=0\to \text{divisible by all positive integers}$

Substitute n=2 in $2^{3n}-7n-1$, we get
$2^6-14-1=49$
$LetP\left(n\right):2^{3n}-7n-1\text{is divisible by 49}$
P(2) is true.
Assume P(k) is true
$2^{3k}-7k-1=49m$
Substituting k+1 in place of n, we get
$2^{3k+3}-7(k+1)-1={8.2}^{3k}-7k-8=8.(2^{3k}-7k-1)+7.7k=49(8m+k)\to \text{divisible by 49}$
P(k+1) is true

Hence, P(n) is true.

Substituting n=1, we have
${10}^1+1=11LetP\left(n\right):{10}^{2n-1}+1\text{is divisible by 11}$

P(1) is true.
Assume P(k) is true
$\Rightarrow {10}^{2k-1}+1=11mNow,P(k+1):{10}^{2k+1}+1={100.10}^{2k-1}+1={99.10}^{2k-1}+{10}^{2k-1}+111.({9.10}^{2k-1}+m)\to \text{divisible by 11}$

P(k+1) is also true.
Hence, P(n) is true. 

Substituting n=1, we get
$7^1-3^1=4$
Let $P\left(n\right):7^n-3^n\text{is divisible by 4}$
P(1) is true
Assume P(k) is true
$7^k-3^k\text{is divisible by 4}{7}^k-3^k=4mNow,7^{k+1}-3^{k+1}=7.(7^k-3^k)+{4.3}^k=4.(7m+3^k)\to \text{divisible by 4}$
P(k+1) is true

Hence, P(n) is true.

Suppose there is a given statement P(n) involving natural numbers n such that

i) The statement is true for a specific natural number m i.e. P(m) is true. This is known as the base case.

ii) If the statement is true for n = k (where k is some positive integer), then the statement is also true for n=k+1, i.e., the truth of P(k) implies the truth of P(k+1).

The second step is called the inductive step.