Find the Sum of Series up to 11 Terms

[rohanprabhu]

Homework Statement

Find the sum of the series:

<br /> S = 1~+~\frac{1 + 2^3}{1 + 2}~+~\frac{1 + 2^3 + 3^3}{1 + 2 + 3}~+~\frac{1 + 2^3 + 3^3 + 4^3}{1 + 2 + 3 + 4}~+~...<br />

upto 11 terms

Homework Equations

Sum of first 'n' natural numbers: S = \frac{n(n + 1)}{2}
Sum of the squares of the first 'n' natural numbers: S = \frac{n(n + 1)(2n + 1)}{6}
Sum of the cubes of the first 'n' natural numbers: S = \left(\frac{n(n+1)}{2}\right)^2

The Attempt at a Solution

In the series, the n^{th} term is given by:

<br /> T_n = \frac{1 + 2^3 + 3^3 + ... + n^3}{1 + 2 + 3 + ... + n}<br />

<br /> T_n = \frac{\left(\frac{n(n+1)}{2}\right)^2}{\left(\frac{n(n + 1)}{2}\right)}<br />

<br /> T_n = \frac{1}{2}(n^2 + n)<br />

Hence,

<br /> S_n = \frac{1}{2}\left(\frac{n(n + 1)(2n + 1)}{6} + \frac{n(n + 1)}{2}\right)<br />

On substituting n = 11, I get:

<br /> S_n = 286<br />

But this isn't one of the options I got on the test. All the options were between 300-400 and all had their last digit '9' [that's all i remember.. they took the question paper away].

 

[Gib Z]

Well from what I see there are no errors. Are you sure you have the question right? Even when we sum 12 terms, we get 364, and 13 would be more than 400.

 

[rohanprabhu]

nope.. the question is definitely this. I remember it since i tried to solve for it quite some time. And nope.. not even 364 is an option...