p(n) 2³ⁿ - 1 is divisible by 8
For n = 1, we get
P(1) = 3^2.1-1 = 9 - 1 = 8
P(1) = 8, which is divisible by 8.
Let P(n) be true for n = k
P(k)3^2k - 1 is divisible by 8......(1)
Now,  P(K+1) = 3^{(2k + 2)} - 1 = 3^2k.3² - 1
= 3² (3^2k - 1) + 8
Now,3^2k-1 is divisible by 9. [Using (1)]
\(\therefore\) 3² (3^2k - 1) + 8 is also divisible by 8.
Hence, 3²ⁿ - 1 is divisible by 8\(\forall\) n \(\in\) N

We have to select at most 3 girls. So the committee consists of no girl and 7 boys or 1 girl and 6 boys or 2 girls and 5 boys or 3 girls and 4 boys.
\(\therefore\) Number of ways of selection ⁴C_o \(\times\) ⁹C₁ + ⁴C₁ \(\times\) ⁹C₆ + ⁴C₂ \(\times\) ⁹C₅ + ⁴C₃ \(\times\) ⁹C₄
\(=1\times {9!\over7!2!}+{4!\over1!3!}+{9!\over6!3!}+{4!\over2!2!}\times{9!\over5!4!}+{4!\over3!1!}+{9!\over4!5!}\)
\(=1\times{9\times8\times7!\over7!\times2\times1}+{4\times3!\over1\times3!}\times{9\times8\times7\times6!\over6!\times3\times2\times1}+{4\times3\times2!\over2\times1\times2!}\times{9\times8\times7\times6\times5!\over 5!\times4\times3\times2\times1} +{4\times3!\over3!\times1}\times{9\times8\times7\times6\times5!\over4\times3\times2\times1\times5!}\)
= 36 + 336 + 756 + 504 = 1632

There are 9 boys and 4 girls. We have to select exactly 3 girls out of 4 girls and 4 boys out of 9 boys.
\(\therefore\) Number of ways of selection = \(^9C_4\times ^4C_3={9!\over 4!5!}\times{4!\over3!\times1!}\)
\(={9\times8\times7\times6\times5\over 4!5!}\times{4!\over 3\times2\times1}=504\)

Total number of points = 18
Out of 18 numbers, 5 are collinear and we get a straight line by joining any two points.
\(\therefore\) Total number of straight line formed by joining 2 points out of 18 points = ¹⁸C₂
Number of straight lines formed by joining 2 points out of 5 points = ⁵C₂
But 5 points are collinear and we get only one line when they are joined pairwise.
So, the required number of straight lines are
=¹⁸C₂ -⁵C₂ +1 = \({18 ·17\over2.1}-{5·4\over2.1}+1= 153 -10 + 1-144\)
Hence, the total number of straight lines = 144

LHS = n!(n + 2)
RHS = n! + (n+1)!
= n!+(n+1)(n)!....(1)
= n!(1+n+1)
= n!(n+20)...(2)
From (1) and (2), LHS = RHS
Hence Proved.