$P\left(1\right):1^2+1\text{is odd}\to nottrueP\left(2\right):2^2+2\text{is odd}\to nottrue$
Suppose P(k) is true.
$k^2+k\text{is odd}\Rightarrow k^2+k=2m+1P(k+1):(k+1{)}^2+(k+1)=(k^2+2k+1)+(k+1)=(k^2+k)+(2k+2)=2m+1+2k+22(m+k+1)+1\Rightarrow \text{P(k+1) is true}$

Check through option, the condition

 ${\left(\frac{n+1}{2}\right)}^n$ ≥ n ! is true for n  ≥ 1.

$P\left(n\right):5^{2n+1}+3^{n+2}{.2}^{n-1}$
$P\left(1\right):5^3+3^3.1=152\to \text{divisible by 19}$
Assume P(k) is true.
$5^{2k+1}+3^{k+2}{.2}^{k-1}\text{is divisible by 19}{5}^{2k+1}+3^{k+2}{.2}^{k-1}=19mP(k+1):5^{2k+3}+3^{k+3}{.2}^k={25.5}^{2k+1}+{6.3}^{k+2}{.2}^{k-1}=6.(5^{2k+1}+3^{k+2}{.2}^{k-1})+{19.5}^{2k+1}=19(6m+5^{2k+1})$

$P\left(n\right):1+3+3^2+...+3^{n-1}=\frac{3^n-1}{2}P\left(1\right):1=\frac{3-1}{2}\Rightarrow 1=1-true$
Assume P(k) is true
$\Rightarrow 1+3+3^2+...+3^{k-1}=\frac{3^k-1}{2}Now,P(k+1):1+3+3^2+...+3^{k-1}+3^k=\frac{3^k-1}{2}+3^k=3^k(1+\frac{1}{2})-\frac{1}{2}=\frac{3^{k+1}-1}{2}$
P(k+1) is also true.
Hence, P(n) is true for all natural numbers

For the proof by mathematical induction to work, the statement P(n) must be true for a specific instance of a natural number.
Hence, if P(m) is true, where m is a specific natural number and P(k+1) is true if P(k) is true for an arbitrary natural number k, then, $P\left(n\right)istrue\forall n\ge m$
Without the base case P(m), we cannot say that P(n) is true. Hence, the statement is false.

The principle of mathematical induction can be used to prove true statements P(n) about natural numbers if it is true for a specific natural number, 'm.' In this case, if proved using induction, P(n) will be true for all $n\ge m.$ 'm' need not be equal to 1.

$P\left(n\right):1+3+5+...+2n-1=n^{2}P\left(1\right):1=1^2\Rightarrow 1=1-true$
Assume P(k) is true
$1+3+5+...+2k-1=k^{2}Now,P(k+1):1+3+5+...+2n-1+2k+1=k^2+2k+1=(k+1{)}^2$
P(k+1) is also true.
Hence, P(n) is true for all natural numbers

$P\left(n\right):1.2+2.3+3.4+...+n(n+1)=\frac{(n+1)(n+2)}{3}P\left(1\right):1.2=\frac{2.3}{3}\Rightarrow 2=2-true$
Assume P(k) is true
$\Rightarrow 1.2+2.3+3.4+...+k(k+1)=\frac{(k+1)(k+2)}{3}Now,P(k+1):1.2+2.3+3.4+...+k(k+1)+(k+1)(k+2)=\frac{(k+1)(k+2)}{3}+(k+1)(k+2)=\frac{4(k+1)(k+2)}{3}\ne \frac{(k+2)(k+3)}{3}$
P(k+1) is not true.
Hence, P(n) is not true for all natural numbers

$P\left(n\right):a^{2n}-b^{2n}\text{is divisible by}a+b,\forall nϵN$

Since we need to prove P(n) for all natural numbers, the base case will be n=1.
Substituting n=1, we get
$a^2-b^2=(a+b)(a-b)\text{is divisible by}a+b$