JEE Main 2022 (Online) 27th July Evening ShiftQuestion 1
$\frac{2^3-1^3}{1\times7}+\frac{4^3-3^3+2^3-1^3}{2\times11}+\frac{6^3-5^3+4^3-3^3+2^3-1^3}{3\times15}+\cdots+\frac{30^3-29^3+28^3-27^3+\ldots+2^3-1^3}{15\times63}$ is equal to ________.
Sequence and Series · Standard Sums (Σ formulas)
Answer$120$
Solution
Formulas used. (1) $(2j)^3-(2j-1)^3=12j^2-6j+1$(2) $\sum_{j=1}^{k}j^2=\frac{k(k+1)(2k+1)}{6}$, $\ \sum_{j=1}^{k}j=\frac{k(k+1)}{2}$Step 1 - The $k$-th numerator. It is $\sum_{j=1}^{k}\left[(2j)^3-(2j-1)^3\right]$: $$\sum_{j=1}^{k}\left(12j^2-6j+1\right)=2k(k+1)(2k+1)-3k(k+1)+k=k\left(4k^2+3k\right)=k^2(4k+3).$$ Step 2 - The $k$-th denominator. $1\times7,\ 2\times11,\ 3\times15,\dots$ is $k(4k+3)$ (and $15\times63$ fits $k=15$). Step 3 - Each term. $$\frac{k^2(4k+3)}{k(4k+3)}=k .$$ Step 4 - Sum. $$1+2+\cdots+15=120 .$$ Answer: 120