using PMI show that 9^(n+1) - 8n - 9 is divisible by 64 whenever n is +ve integer
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➤ 9^(n+1) - 8n - 9
=9ⁿ×9 - 8n - 9
= (1 + 8)ⁿ×9 - 8n - 9
Now, break (1+8)ⁿ by using binomial formula,
(x + y)^n = nC0.x^n + nC1.x^(n-1)y + nC2.x^(n-2)y^2 + ........+nCn.y^n
(1+8)^n = nC0 + nC1(8) + nC2(8)² +....... + nCn(8)ⁿ use this above
= {nC0 + nC1.8 + nC2.8² + .....+ nCn.8ⁿ}×9 - 8n - 9
= (1 + 8n + nC2.8² + ......+nCn.8ⁿ)×9 - 8n - 9
= 9 + 72n + (nC2.8² +nC3.8³ +......+ nCn.8ⁿ)9 - 8n - 9
= (72n -8n) + 8²[nC2 + nC3.8 +.....+ nCn.8^(n-2)]
= 64n + 64[nC2 + nC3.8 +.....+nCn.8^(n-2)]
= 64 ×constant number
we obeserved that,
9^(n+1) - 8n -9 is divisible by 64
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