See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

  n k n k n 2k k 0 S 4 C        n 1 k n k 1 n 1 2k k 0 S 4 C          n 1 k n k 1 n 1 2k k 0 S 4 C        Consider n 1 n 1 n r r r C C 2 C      coefficient of xr in {(1 + x)n+1 + (1 + x)n – 1 - 2(1 + x)n} = coefficient of xr in (1 + x)n–1 {(1 + x)2 + 1 – 2(1 + x)} = coefficient of xr in (1 + x)n–1 {x + 2x + 2 – 2x – 2} = coefficient of xr–2 in (1 + x)n–1 = n– 1 Cr – 2 Put n n + k and r  2k   n k 1 n k 1 n k n k 1 2k 2k 2k 2 k 1 C C 2 C C .             Multiplying by (–4)k and applying summation on the both sides           n 1 n 1 n n 1 k k k k n k 1 n k 1 n k n k 1 2k 2k 2k 2 k 1 k 0 k 0 k 0 k 1 4 C 4 C 2 4 C 4 C                        (Note that the limits of summation are different for each term)     n 1 k n k 1 n 1 n 1 n 2 k 1 k 1 S S 2S 4 C             Let k – 1 = t    n t 1 n t 2t n t 0 4 4 C 4S       n 1 2 n 1 S 2S S 0      Put n = 2010.