Let A be the sum of the first 20 terms and B be the sum of the first 40 terms of the series $$1^2 + 2 \cdot 2^2 + 3^2 + 2 \cdot 4^2 + 5^2 + 2 \cdot 6^2 + \ldots$$ If $$B - 2A = 100\lambda$$, then $$\lambda$$ is equal to:

We look carefully at the given series

$$1^{2}+2\cdot2^{2}+3^{2}+2\cdot4^{2}+5^{2}+2\cdot6^{2}+\ldots$$

and notice a very clear pattern: for every odd natural number its square appears once, while for every even natural number its square appears twice. Thus, for the general term with index $$n$$ we may write

$$\text{term}(n)=\begin{cases}1\cdot n^{2}, & n\text{ odd},\\[4pt]2\cdot n^{2}, & n\text{ even}.\end{cases}$$

Let us denote by $$c_{n}$$ this coefficient (1 for odd, 2 for even), so that the partial sum of the first $$k$$ terms is

$$S(k)=\sum_{n=1}^{k}c_{n}\,n^{2}.$$

We are told that

$$A=S(20),\qquad B=S(40).$$

To evaluate these sums we separate the odd and even indices.

For any positive integer $$k$$ we write

$$S(k)=\sum_{\substack{n=1\\n\text{ odd}}}^{k}n^{2}+\;2\sum_{\substack{n=1\\n\text{ even}}}^{k}n^{2}.$$

Now we change the running index:

• Put $$n=2m-1$$ for the odds. When $$n$$ runs through the first $$k$$ integers, $$m$$ runs from $$1$$ to $$\left\lceil\dfrac{k}{2}\right\rceil.$$
• Put $$n=2m$$ for the evens. Then $$m$$ runs from $$1$$ to $$\left\lfloor\dfrac{k}{2}\right\rfloor.$$

Hence, with $$o=\left\lceil\dfrac{k}{2}\right\rceil$$ (number of odds) and $$e=\left\lfloor\dfrac{k}{2}\right\rfloor$$ (number of evens), we get

$$S(k)=\sum_{m=1}^{o}(2m-1)^{2}+2\sum_{m=1}^{e}(2m)^{2}.$$

We require explicit formulas for the two kinds of sums. We therefore recall the standard identities

1. Sum of first $$n$$ natural numbers:

$$\sum_{m=1}^{n} m = \dfrac{n(n+1)}{2}.$$

2. Sum of squares of first $$n$$ natural numbers:

$$\sum_{m=1}^{n} m^{2} = \dfrac{n(n+1)(2n+1)}{6}.$$

Using these two, we evaluate:

$$\sum_{m=1}^{n}(2m)^{2}=4\sum_{m=1}^{n}m^{2}=4\cdot\dfrac{n(n+1)(2n+1)}{6}.$$

For the odd squares we expand

$$(2m-1)^{2}=4m^{2}-4m+1,$$

so that

$$\sum_{m=1}^{n}(2m-1)^{2}=4\sum_{m=1}^{n}m^{2}-4\sum_{m=1}^{n}m+\sum_{m=1}^{n}1 =4\left(\dfrac{n(n+1)(2n+1)}{6}\right)-4\left(\dfrac{n(n+1)}{2}\right)+n.$$

We are now ready to compute $$A$$ and $$B$$ explicitly.

Computation of $$A=S(20)$$

The first 20 integers comprise 10 odds and 10 evens, so $$o=e=10.$$

First, for $$n=10$$ we have

$$\sum_{m=1}^{10}m^{2}=\dfrac{10\cdot11\cdot21}{6}=385,$$

$$\sum_{m=1}^{10}m=\dfrac{10\cdot11}{2}=55.$$

Therefore

$$\sum_{m=1}^{10}(2m-1)^{2}=4(385)-4(55)+10=1540-220+10=1330,$$

and

$$2\sum_{m=1}^{10}(2m)^{2}=2\left[4\sum_{m=1}^{10}m^{2}\right]=2(4\cdot385)=2(1540)=3080.$$

Adding the two pieces we find

$$A=1330+3080=4410.$$

Computation of $$B=S(40)$$

For the first 40 integers we have 20 odds and 20 evens, so $$o=e=20.$$

For $$n=20$$ we obtain

$$\sum_{m=1}^{20}m^{2}=\dfrac{20\cdot21\cdot41}{6}=2870,$$

$$\sum_{m=1}^{20}m=\dfrac{20\cdot21}{2}=210.$$

Thus

$$\sum_{m=1}^{20}(2m-1)^{2}=4(2870)-4(210)+20=11480-840+20=10660,$$

and

$$2\sum_{m=1}^{20}(2m)^{2}=2\left[4\sum_{m=1}^{20}m^{2}\right]=2(4\cdot2870)=2(11480)=22960.$$

Whence

$$B=10660+22960=33620.$$

Finding $$B-2A$$

We now calculate

$$2A=2(4410)=8820,$$

so

$$B-2A=33620-8820=24800.$$

The statement of the problem tells us that

$$B-2A=100\lambda,$$

which gives directly

$$\lambda=\dfrac{24800}{100}=248.$$

Hence, the correct answer is Option C.