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) : 3^{2n + 2} - 8ⁿ - 9 is divisible by 8..
Then, P(1): 3^2.1+2 - 8.1 - 9 is divisible by 8.
(i.e.) 3⁴ - 8 - 9 is divisible by 8 or 81 - 8 - 9 is divisible by 8
(or) 64 is divisible by 8, which is true.
Suppose P(k) is true, then
P(k) : 3^{2k + 2} - 8k - 9 is divisible by 8
(i.e.) 3^2k+2-8k-9 = 8m,where m\(\in\)N (or)
3^2k+2 = 8m + 8k+ 9
P(k + 1) is the statement given by, ...(1)
P(k + 1): 3^{2(k+ 1)+2} - 8(k-1) - 9
Now, 3^{2(k+ 1)+2} - 8(k- 1) - 9 = 3^{2k+ 4} -8k - 8 - 9 = 3^{2k+ 4} - 8k - 17
= 9.3^{2k+ 2} - 8k- 17
9(8m + 8k + 9) - 8k - 17 [ \(\therefore\) P(k) is true. Using (1)]
\(\therefore\) 3^{2(k+ 1)+2} - 8(k- 1) - 9 is divisible by 8.
\(\therefore\) P(k + 1) is true.
Hence, by the principle of mathematical induction, P(n) is true for all n \(\in\) N

There are 9 women and 8 men. A committee of 12, consisting of atleast 5 women can be formed by the following cases:

\(\therefore\) Total number of ways of forming the committee
= 128 \(\times\) 8 + 84 \(\times\) 28 + 36 \(\times\) 56+ 9 \(\times\) 70 + 1 \(\times\) 56 = 6062.
Clearly, women are majority in cases (c), (d) and (e).
\(\therefore\) Total number of committees in which women are majority = 9C₇ \(\times\) 8C₅ \(\times\) 9C₈ \(\times\) 8C₄ \(\times\) 9C₉ \(\times\) 8C₃
= 36 \(\times\) 56 + 9 \(\times\) 70 + 1\(\times\)56
= 2702

There are 5 vowels in the given word. These vowels can be put together in 5! ways.
Considering these 5 vowels as one letters we have 8 letters of which 2 are T's.
These 8 letters can be arranged in \(\frac{8!}{2!}\) ways.
Hence, by the fundamental principle of multiplication, required number of words
= 5! \(\times\) \(\frac{8!}{2!}\) = 5! \(\times\) \(\frac { 8\times 7\times 6\times 5\times 4\times 3\times 2! }{ 2! } \)
= 5 \(\times\) 4 \(\times\) 3 \(\times\) 2 \(\times\) 8 \(\times\) 7 \(\times\) 6 \(\times\) 5 \(\times\) 4 \(\times\) 3 = 2419200

Given 2_n+1 P_n-1: 2_n-1 P_n = 3 : 5
\(⇒\ {2n+1P_{n-1}\over2n-2P_n}={3\over 5}\)
⇒ 5· (2_n+1)P_n-1 = 3· 2_n-1 P_n​
\(⇒\ \ 5\times{(2n+1)!\over (2n+1-n+1)!}=3.{(2b-1)!\over (2n-1-n)!}\)

\(⇒\ \ {5\times (2n+1)(2n)\over (n+2)(n+1)(n)(n-1)!}={3\over (n-1)!}\)
\(⇒\ \ {10(2n+1)\over (n+2)(n+1)}=3\)
⇒ 10(2n + 1) = 3(n + 2)(n + 1)
⇒ 20n+ 10=3(n²+3n+2)
20n + 10 - 3n2 - 9n - 6 = 0
⇒ -3n² + 11n + 4 = 0
⇒ 3n² - 11n - 4 = 0
⇒ \((n-4)(3n+1)=0\Rightarrow n=4\)