Find the Sum of Series up to 11 Terms

[rohanprabhu]

Homework Statement

Find the sum of the series:

$$S = 1~+~\frac{1 + 2^3}{1 + 2}~+~\frac{1 + 2^3 + 3^3}{1 + 2 + 3}~+~\frac{1 + 2^3 + 3^3 + 4^3}{1 + 2 + 3 + 4}~+~...$$

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:

$$T_n = \frac{1 + 2^3 + 3^3 + ... + n^3}{1 + 2 + 3 + ... + n}$$

$$T_n = \frac{\left(\frac{n(n+1)}{2}\right)^2}{\left(\frac{n(n + 1)}{2}\right)}$$

$$T_n = \frac{1}{2}(n^2 + n)$$

Hence,

$$S_n = \frac{1}{2}\left(\frac{n(n + 1)(2n + 1)}{6} + \frac{n(n + 1)}{2}\right)$$

On substituting n = 11, I get:

$$S_n = 286$$

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...