JEE Main 2020 (Online) 7th January Morning ShiftQuestion 1
If the sum of the coefficients of all even powers of $x$ in the product $\left(1+x+x^{2}+\ldots+x^{2n}\right)$ $\left(1-x+x^{2}-x^{3}+\ldots+x^{2n}\right)$ is $61$, then $n$ is equal to
Binomial Theorem · Properties and Sums of Binomial Coefficients
Answer$30$
Solution
Concept. The even-power coefficients add up to $\frac{f(1)+f(-1)}{2}$. Formulas used. (1) $\sum_{\text{even}}a_j=\dfrac{f(1)+f(-1)}{2}$Step 1 - $f(1)$. The first factor has $2n+1$ terms, all $1$: value $2n+1$. The second alternates $1-1+\cdots+1$ with $2n+1$ terms: value $1$. So $f(1)=2n+1$. Step 2 - $f(-1)$. Now the first factor alternates (value $1$) and the second has all terms $+1$ (value $2n+1$). So $f(-1)=2n+1$. Step 3 - Solve. By (1), $\dfrac{(2n+1)+(2n+1)}{2}=2n+1=61$, so $n=30$. Answer: 30