JEE Main 2019 (Online) 8th April Morning ShiftQuestion 1
The sum of the series $2\cdot{}^{20}C_0+5\cdot{}^{20}C_1+8\cdot{}^{20}C_2+11\cdot{}^{20}C_3+\ldots+62\cdot{}^{20}C_{20}$ is equal to
Binomial Theorem · Properties and Sums of Binomial Coefficients
Solution
Concept. The multipliers $2,5,8,\dots,62$ are $3r+2$; split the sum. Formulas used. (1) $\sum{}^{n}C_r=2^n$ and $\sum r\,{}^{n}C_r=n\,2^{n-1}$Step 1 - General term. $(3r+2)\,{}^{20}C_r$ for $r=0,\dots,20$ (last: $3\cdot20+2=62$). Step 2 - Sum. By (1), $$3\cdot20\cdot2^{19}+2\cdot2^{20}=60\cdot2^{19}+4\cdot2^{19}=64\cdot2^{19}=2^{25}.$$ Answer: (B)