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

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

1. ## ⭐Question:–

### Solution:-

Let p (x) =

[tex]3 {}^{2n + 2} \: – 8x \: – \: 9 \: is \: divisble \: by \: 64 \\ [/tex]

[tex]when \: put \: n \: = \: 1 \\ [/tex]

[tex]p \: (1) = \: 3 {}^{4} \: – 8 \: – 9 = \: 64[/tex]

Let n = k and we get,

[tex]p \: (k) \: = \: 3 {}^{2k + 2} \: – \: 8k \: – 9 \: divisble \: by \: 64[/tex]

[tex]3 {}^{2k + 2} \: – 8k \: – 9 \: = \: 64m \: \: \: \: when \: m \: =N\\ [/tex]

[tex]now \: we \: shall \: prove \: that \: p {}^{k + 1} is \: also \: true[/tex]

[tex]p \: (k + 1) \: = \: 3 {}^{2(k + 1) + 2} \: – 8(k + 1) – 9 \: is \\ divisble \: by \: 64[/tex]

Now,

[tex]p(k + 1) = \: 3 {}^{2(k + 1) + 2} – 8(k + 1) \: – 9 = 3 {}^{2} .3 {}^{2k + 2} – 8k \: – 17[/tex]

[tex] = \: 9.3 {}^{2k + 2} – 8k – 17[/tex]

[tex] = \: 9(64m + 8k + 9) \: – 8k \: – 17[/tex]

[tex] = 9.64m \: 72k + 81 – 8k – 17[/tex]

[tex] = 9.64m + 64k + 64[/tex]

[tex] = 64(9m + k + 1)[/tex] which is disability by 64.