# Sum of series

1. Apr 20, 2008

### rohanprabhu

1. The problem statement, all variables and given/known data

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

2. Relevant 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: $$S = \left(\frac{n(n+1)}{2}\right)^2$$

3. 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].

Last edited: Apr 20, 2008
2. Apr 20, 2008

### 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.

3. Apr 20, 2008

### rohanprabhu

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