JEE Main 2022 · 29 Jun · Shift 2 · Mathematics Q20

MathematicsSingle correct+4 / -1Question 20 of 90

Let $n\ge5$ be an integer. If $9^n-8n-1=64\alpha$ and $6^n-5n-1=25\beta$ , then $\alpha-\beta$ is equal to

1+{}^nC_2(8-5)+{}^nC_3\left(8^2-5^2\right)+\dots+{}^nC_n\left(8^{n-1}-5^{n-1}\right)

1+{}^nC_3(8-5)+{}^nC_4\left(8^2-5^2\right)+\dots+{}^nC_n\left(8^{n-2}-5^{n-2}\right)

{}^nC_3(8-5)+{}^nC_4\left(8^2-5^2\right)+\dots+{}^nC_n\left(8^{n-2}-5^{n-2}\right)

{}^nC_4(8-5)+{}^nC_5\left(8^2-5^2\right)+\dots+{}^nC_n\left(8^{n-3}-5^{n-3}\right)

Answer and worked solution

Official answer: C

Expand. $9^n-8n-1=\sum_{k\ge2}{}^nC_k8^k$ , so $\alpha=\sum_{k\ge2}{}^nC_k8^{k-2}$ ; likewise $\beta=\sum_{k\ge2}{}^nC_k5^{k-2}$ .

Difference. The $k=2$ terms cancel: $\alpha-\beta=\sum_{k\ge3}{}^nC_k\left(8^{k-2}-5^{k-2}\right)$ .

(C)

Topics:Binomial expansion

JEE Main 2022 · 29 Jun · Shift 2, Mathematics question 20