Let p(n) : 1 + 5 + 9 + ... + (4n - 3) = n(2n - 1), \(\forall\)n\(\in\)N
Step 1: P(1) 1 = 1(2.1 - 1) = 1 which is true for P(1)
Step 2: P(k)1 + 5 + 9 + ... + (4k- 3) = k(2k -1). Let it be true.
Step 3: P(k): 1 + 5 + 9 + ... + (4k- 3) = k(4k+ 1)
= k(2k -1) + (4k + 1) = 2k2 - k + 4k + 1
= 2k2 + 3k + 1 = 2k² + 2k + k + 1
= 2k(k+ 1) + 1 (k+ 1) = (2k+ 1)(k+ 1)
= (k+ 1) (2k+ 2 -1) = (k+ 1) [2(k+ 1) -1]
Which is true for P(k + 1).
Hence, P(k + 1) is true whenever P(k) is true

Let P(n) : n³ - n
Step 1: P(2) : 2³ - 2 = 6 which is divisible by 6. So it is true for P(2).
Step 2 : P(k) :k³ - k = 6\(\lambda\)Let it is be true for k\(\ge\) 2
\(\Rightarrow\) k³ = 6\(\lambda\)+ k ...(i)
Step 3 : P(k+ 1) = (k+ 1)³-(k+ 1)
= k³ ​​​​​​+1+ 3k²+ 3k - k - 1
= k³ + k + 3(k²+ k)
= 6\(\lambda\)+ 3(k² + k) [from (i)]
We know that 3(k² + k) is divisible by 6 for every value of k\(\in\)N.
Hence P(k + 1) is true whenever P(k) is true.

Let p(n) : 2 + 4 + 6 + ...+ 2n = n² + n, \(\forall\) n \(\in\) N
Step 1: P(1): 2 = 1² + 1 = 2 which is true for P(1)
Step 2: P(k) : 2 + 4 + 6 + ...+ 2k = k² + k. Let it be true.
Step 3: P(k+ 1): 2 + 4 + 6 + ...+ 2k+ 2k+ 2
= k² + k + 2k + 2 = k²+ 3k + 2
= k²+ 2k + k + 1 + 1
= (k+ 1)² + (k+ 1)
Which is true for P(k + 1)
So, P(k + 1) is true whenever P(k) is true.

Suppose a² = x and \(\sqrt{a^{2}-1}=y\)
∴ \(a^{2}+\sqrt{a+1}=x+y \) and \(a^{2}-\sqrt{a^{2}-1}=x-y\).
Using binomial theorem,
\((x+y)^{4}={^4C_{0}x^{4}+{^4C_{2}x^{3}y}+{^4C_{2}x^{2}y^{2}}+{^4C_{3}xy^{3}+{^4C_{4}y^{4}}}}\)
⇒ (x+y)⁴ = x⁴+ 4x³y + 6x²y² + 4xy³ + y⁴ and
(x-y)⁴ = ⁴C₀x⁴ - ⁴C₁x³y + ⁴C₂x²y² - ⁴C₃xy³ + ⁴C₄y⁴
⇒ \((x-y)^{4}=x^{4}-4x^{3}y+6x^{2}y^{2}-4xy^{3}+y^{4}\)
∴ \((x+y)^{4}+(x-y)^{4}=2[x^{4}+6x^{2}y^{2}+y^{4}]\)
\(\therefore (a^{2}+\sqrt{a^{2}+1})^{4}+(a^{2}-\sqrt{a^{2}-1})^{4}=2[(a^{2})^{4}+6(a^{2})^{2}(\sqrt{a^{2}-1})^{2}+(\sqrt{a^{2}-1})^{4}] \)
= \(2[a^{8}+6a^{4}(a^{2}-1)+(a^{2}-1)^{2}]\)
= \(2[a^{8}+6a^{6}-6a^{4}+a^{4}-2a^{2}+1]\)
= \(2[a^{8}+6a^{6}-5a^{4}-2a^{2}+1]\)

\(\frac{2^n+3^n}{n}\)

Let 1.3 + 2.3² + 3.3³+...+n-3ⁿ
\(={(2-n1)3^{n+1}+3\over4}\)
Step 1:
\(p(1)=1.3={(2\times1-1).3^2+3\over 4}\)
\(3={9+3\over 4}=3\)
p(1) is true
Step 2:
Let p(m) be true. Then
\(1.3+2.3^2+...+m.3^m={}(2m-1)3^{m+1}+3\over 4\)
Step 3:
To prove that p(m+1) is true
i.e To prove that \(1.3+2.3^2+...+m.3^m+(m+1)3^{m+1}={[2(m+1)-1]3^{m+2}+3\over 4}\)
\(={(2m+1).3^{m+2}+3\over 4}\)
Consider 1.3+2.3² + ...+m.3^m+(m+1)3^m+1
\(={(2m-1)3^{m+1}+3\over 4}+(m+1)3^{m+1}\)
\(={(2m-1)3^{m+1}+3+4(m+1)3^{m+1}\over 4}\)
\(={(2m-1)3^{m+1}+(4m+4)3^{m+1}+3\over 4}={3^{m+1}[2m-1+4m+4]+3\over 4}\)
\(={3^{m+1}(6m+3)+3\over 4}\)
\(={3^{m+1}3(2m+1)+3\over4}\)
\(={3^{m+2}(2m+1)+3\over4}=RHS\)
⇒ p(m+1) is true
Hence, by the principle of mathematical induction, p(n) is true for all n ∈ N

In a pack of 52 cards, there are four aces
3 aces will be selected from 4 cards and remaining 2 cards will be selected from rest of 48 cards.
∴ Number of ways of selecting 3 aces from 4 aces = 4C₃
Number of ways of selecting 2 cards out of 48 cards = 48C₂
∴ Required number of ways = 4C₃ \(\times \) 48C₂
= \(4\times \frac { 48\times 47 }{ 2\times 1 } \)     [∵ nC_r = 1C_n-r]
= \(2\times 48\times 47\)
= 4512

Let T_n be the nth term of the given series.
\(\therefore { T }_{ n }=\frac { { 1 }^{ 3 }+{ 2 }^{ 3 }+3^{ 3 }+....+n\quad terms }{ 1+3+5....+n\quad terms } =\frac { \Sigma { n }^{ 3 } }{ \frac { n }{ 2 } (2(1)+(n-1)2) } \)
\(\frac { { \left( \frac { n(n+1) }{ 2 } \right) }^{ 2 } }{ \frac { n }{ 2 } (2n) } \left[ \because \Sigma { n }^{ 3 }={ \left( \frac { n(n+1) }{ 2 } \right) }^{ 2 } \right] \)
\(=\frac { { n(n+1) }^{ 2 } }{ \frac { n }{ 2 } (2n) } =\frac { 1 }{ 4 } ({ n }^{ 2 }+2n+1)\)
\(=\frac { { n(n+1) }^{ 2 } }{ \frac { n }{ 2 } (2n) } =\frac { 1 }{ 4 } ({ n }^{ 2 }+2n+1)\)
\(=\frac { 1 }{ 4 } (\Sigma { n }^{ 2 }+\Sigma 2n+n)=\frac { 1 }{ 4 } \left[ \frac { n(n+1)(2n+1) }{ 6 } +\frac { 2.n(n+1) }{ 2 } +n \right] \)
\(=\frac { 1 }{ 4 } \left[ \frac { n(n+1)(2n+1)+6n(n+1)+6n }{ 6 } \right] \)
\(=\frac { n }{ 24 } \left[ { 2n }^{ 2 }+9n+13 \right] =\frac { n({ 2n }^{ 2 }+9n+13) }{ 24 } \)

LHS = ²ⁿC_n
= \(\frac { 2n! }{ n!(2n-n)! } =\frac { 2n! }{ n!n! } \)
= \(\frac { (2n)(2n-1)(2n-2)...4.3.2.1 }{ n!n! } =\frac { (2n)(2n-2)...4.2[(2n-1)(2n-3)...3.1] }{ n!n! } \)    [Separate the even and odd terms]
= \(\frac { { 2 }^{ n }.n(n-1)...2.1[(2n-1)(2n-3)...3.1] }{ n!n! } \)
= \(\frac { { 2 }^{ n }.(n)![1.3.5...(2n-3)(2n-1)] }{ n!n! } =\frac { { 2 }^{ n }[1.3.5...(2n-3)(2n-1)] }{ n! } \) = RHS

If we take y = 3x, then \({1\over (1+3x)^2}={1\over (1+y)^2}\)
Now \({1\over(1+y)^2}\) can be expanded using binomial theorem in powers of y. The expansion is valid only for values of y satisfying lyl < 1.
Replacing y by 3x we can get an expansion of \({1\over (1+3x)^2}.\) 
The expansion is valid only for values of x satisfying |3xl< 1; that is the expansion is valid only for values of x satisfying Ixl < \(1\over3\)
\({1\over (1+3x)^2}=(1+3x)^{-2}\)
\(=1-2(3x)+{2(2+1)\over 2!}(3x)^2-{2(2+1)(2+2)\over3!}(3x)^3+{2(2+1)(2+2)(2+3)\over 4!}(3x)^4-......\)
Hence, \({1\over (3+2x)^2}=1-6x+27x^2-108x^3+405x^4-...,|x|<{1\over 3}\)