# Sum of Arithmetic Series help

1. Sep 19, 2010

### DorumonSg

Compute the following sums and give your answers in terms of n.

(a) 1 + 2 + 3 + ... + n
(b) 2 + 4 + 6 + ... + 2n
(c) 1 + 3 + 5 + ... + (2n + 1)

I actually have the answers but there seem to be an error...

For (a) 1 + 2 + 3 + ... + n,

For(b) 2 + 4 + 6 + ... + 2n

However, for (c)

My answer given is (2n + 2)(n + 1)/2

If we play with the answer for (a) and (b) for example :

(a) : We take the nth term as 3 and sub it in n(n+1)/2 = 3(3+1)/2 = 6(Which is the correct sum up to the nth term)

But for (c) :

We take the nth term as 3 and sub it in (2n + 2)(n + 1)/2 = (2(3) + 2)(3 + 1)/2 = 16(Which is the sum of the number AFTER the nth term.)

The answer for (c) is incorrect, am I right? Because its the answer give by my tutor.

2. Sep 19, 2010

### Mentallic

For c) we can manipulate it in such a way as to turn it into a form we easily recognize:

$$1+3+5+...+(2n+1)=(n+1)+0+2+4+...+2n=(n+1)+2(1+2+3+...+n)$$

which gives you $$(n+1)^2$$ as your tutor gave. If you take n=3 then you need to realize that the last term is (2n+1)=7 so you go up until 7... 1+3+5+7=16 as expected.

Last edited: Sep 19, 2010