In order to show that 9^{n+1} – 8n – 9 is divisible by 64, it has to be show that 9^{n+1} – 8n – 9 = 64 k, where k is some natural number

Using binomial theorem,

(1 + a)^{m} = ^{m}C_{0} + ^{m}C_{1} a + ^{m}C_{2} a^{2} + …. + ^{m }C _{m} a^{m}

For a = 8 and m = n + 1 we get

(1 + 8)^{n+1} = ^{n+1}C_{0} + ^{n+1}C_{1} (8) + ^{n+1}C_{2} (8)^{2} + …. + ^{n+1 }C _{n+1 }(8)^{n+1}

9^{n+1} = 1 + (n + 1) 8 + 8^{2} [^{n+1}C_{2} + ^{n+1}C_{3} (8) + …. + ^{n+1 }C _{n+1 }(8)^{n-1}]

9^{n+1} = 9 + 8n + 64 [^{n+1}C_{2} + ^{n+1}C_{3} (8) + …. + ^{n+1 }C _{n+1 }(8)^{n-1}]

9^{n+1} – 8n – 9 = 64 k

Where k = [^{n+1}C_{2} + ^{n+1}C_{3} (8) + …. + ^{n+1 }C _{n+1 }(8)^{n-1}] is a natural number

Thus, 9^{n+1} – 8n – 9 is divisible by 64, whenever n is positive integer.

Hence the proof

Answered by Abhisek | 1 year agoFind the term independent of x in the expansion of \( (\dfrac{3}{2x^2} – \dfrac{1}{3x})^9\)

Find the middle term in the expansion of \((x-\dfrac{ 1}{x})^{2n+1}\)

Find the middle term in the expansion of (1 + 3x + 3x^{2} + x^{3})^{2n}

Find the middle term in the expansion of (1 – 2x + x^{2})^{n}

Find the middle term in the expansion of \( (\dfrac{x}{a} – \dfrac{a}{x})^{10}\)