Let A be the sum of the first 20 terms and B be the sum of the first 40 terms of the serie…

Question

Let $A$ be the sum of the first 20 terms and $B$ be the sum of the first 40 terms of the series $1^2+2\cdot2^2+3^2+2\cdot4^2+5^2+2\cdot6^2+\cdots$. If $B-2A=100\lambda$, then $\lambda$ is equal to:

Accepted Answer

Group the terms in pairs (odd, even): the $i$-th pair is $(2i-1)^2+2(2i)^2 = 4i^2-4i+1+8i^2 = 12i^2-4i+1$.

The sum of the first $m$ pairs is
$$P(m)=\sum_{i=1}^m(12i^2-4i+1)=2m(m+1)(2m+1)-2m(m+1)+m.$$

First 20 terms = 10 pairs, so $A=P(10)$; first 40 terms = 20 pairs, so $B=P(20)$.

$$P(10)=2(10)(11)(21)-2(10)(11)+10 = 4620-220+10=4410.$$
$$P(20)=2(20)(21)(41)-2(20)(21)+20 = 34440-840+20=33620.$$

$$B-2A = 33620-2(4410)=33620-8820=24800 = 100(248)\ \Rightarrow\ \lambda=248.$$

Solutions · 1