JEE Advanced 2021 Paper 2 · Mathematics · Binomial Theorem · hard

Let N=Σk=1^2021kbinom2021k2^k-1. The remainder when N is divided by 7 is

Numerical answer

Let

N=\displaystyle\sum_{k=1}^{2021}k\binom{2021}{k}2^{k-1}

. The remainder when

N

is divided by

7

Show answer and solution

Answer: 6

Use

k\binom{n}{k}=n\binom{n-1}{k-1}

. Then

N=2021\sum_{k=1}^{2021}\binom{2020}{k-1}2^{k-1}=2021\cdot(1+2)^{2020}=2021\cdot3^{2020}

. Now

2021=7\cdot288+5\equiv5\pmod7

. Since

3^6\equiv1\pmod 7

(Fermat) and

2020=6\cdot336+4

, we get

3^{2020}\equiv3^4=81\equiv4\pmod7

. Therefore

N\equiv5\cdot4=20\equiv6\pmod7

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