# SUm

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• Feb 18th 2009, 12:22 AM
thereddevils
SUm
Find the first three terms and the nth term of the series with the sum to n terms , $\displaystyle S_n$ , given by :

(1) $\displaystyle S_n=4n-\frac{1}{n}$

I know the first three terms . The answer given by the book for the nth term is :

$\displaystyle 4+\frac{1}{n(n-1)},n>1$

My question is if that's the answer , why can't i get the first term which is 3 when i substitute n=1 into the formula .

And if it says n is greater than 1 , then there is no way to find the first term.
• Feb 18th 2009, 12:37 AM
CaptainBlack
Quote:

Originally Posted by thereddevils
Find the first three terms and the nth term of the series with the sum to n terms , $\displaystyle S_n$ , given by :

(1) $\displaystyle S_n=4n-\frac{1}{n}$

I know the first three terms . The answer given by the book for the nth term is :

$\displaystyle 4+\frac{1}{n(n-1)},n>1$

My question is if that's the answer , why can't i get the first term which is 3 when i substitute n=1 into the formula .

And if it says n is greater than 1 , then there is no way to find the first term.

The form for the general term $\displaystyle a_n$ only applies to $\displaystyle n>1$ (you need $\displaystyle S_{n-1}$ to get this from the sum to $\displaystyle n$ terms so only applies when $\displaystyle S_{n-1}$ is given by the formula. But $\displaystyle S_0=0$ so the first term:

$\displaystyle a_1=S_1$

CB