If the sum of the first ten terms of the series (135)^2+(225)^2+(315)^2+4^2+(445)^2+ is 16…

Question

If the sum of the first ten terms of the series $\left(1	frac35ight)^2+\left(2	frac25ight)^2+\left(3	frac15ight)^2+4^2+\left(4	frac45ight)^2+\cdots$ is $\dfrac{16}{5}m$, then $m$ is equal to:

Accepted Answer

The $n$-th term (for $n=1,\ldots,10$) is $\left(\dfrac{4(n+1)}{5}\right)^2$, i.e. the numerators run $8,12,16,\ldots,44$ (over $5$), corresponding to $k=2,\ldots,11$ in $\left(\dfrac{4k}{5}\right)^2$.

$$S=\frac{16}{25}\sum_{k=2}^{11}k^2=\frac{16}{25}\left(\frac{11\cdot12\cdot23}{6}-1\right)=\frac{16}{25}(505)=\frac{16}{5}\times101.$$

So $m=101$.

Solutions · 1