Question

Using binomial theorem, prove that 32n+2 - 8n - 9 is divisible by 64.​

Answer 1

⭐Question:-

Using binomial theorem, prove that

$Using \: binomial \: theorem, \: prove \: that\: \\ 3 {}^{2n + 2 \: } - \: 8x \: - \: 9 \: is \: divisble \: by \: 64.$

⭐Answer:-

Solution:-

Let p (x) =

$3 {}^{2n + 2} \: - 8x \: - \: 9 \: is \: divisble \: by \: 64$

$when \: put \: n \: = \: 1$

$p \: (1) = \: 3 {}^{4} \: - 8 \: - 9 = \: 64$

Let n = k and we get,

$p \: (k) \: = \: 3 {}^{2k + 2} \: - \: 8k \: - 9 \: divisble \: by \: 64$

$3 {}^{2k + 2} \: - 8k \: - 9 \: = \: 64m \: \: \: \: when \: m \: =N$

$now \: we \: shall \: prove \: that \: p {}^{k + 1} is \: also \: true$

$p \: (k + 1) \: = \: 3 {}^{2(k + 1) + 2} \: - 8(k + 1) - 9 \: is \\ divisble \: by \: 64$

Now,

$p(k + 1) = \: 3 {}^{2(k + 1) + 2} - 8(k + 1) \: - 9 = 3 {}^{2} .3 {}^{2k + 2} - 8k \: - 17$

$= \: 9.3 {}^{2k + 2} - 8k - 17$

$= \: 9(64m + 8k + 9) \: - 8k \: - 17$

$= 9.64m \: 72k + 81 - 8k - 17$

$= 9.64m + 64k + 64$

$= 64(9m + k + 1)$ which is disability by 64.

Thus, p (k+1) is true whenever p(k) is true.

Hence, by principle mathematical induction,

px is true for all natural number.