JEE Main 2018 (Offline)Question 1
Let $A$ be the sum of the first $20$ terms and $B$ be the sum of the first $40$ terms of the series
$1^2+2\cdot2^2+3^2+2\cdot4^2+5^2+2\cdot6^2+\ldots$
If $B-2A=100\lambda$, then $\lambda$ is equal to :
Sequence and Series · Standard Sums (Σ formulas)
Solution
Formulas used. (1) $\sum_{k=1}^{n}k^2=\frac{n(n+1)(2n+1)}{6}$(2) Sum of even squares: $2^2+4^2+\cdots+(2m)^2=4\sum_{j=1}^{m}j^2$Step 1 - Rewrite. Every term is $k^2$, and the even-place terms get one extra $k^2$: $$\text{Sum of }N\text{ terms}=\sum_{k=1}^{N}k^2+\sum_{\text{even }k\le N}k^2 .$$ Step 2 - $A$ ($N=20$). $$A=2870+4(385)=2870+1540=4410 .$$ Step 3 - $B$ ($N=40$). $$B=22140+4(2870)=22140+11480=33620 .$$ Step 4. $$B-2A=33620-8820=24800=100\times248 .$$ Answer: (C)