# find the kth term from the nth partial sum

Okay this is a very stupid question but i dont know why I dont get it so im sorry in advance

the expression for the nth partial sum of a series $$\sum_{k=1}^\infty u_k$$ is $$s_n = {(3n^2 - 1)}$$

we have to find an expression for $$u_k$$

so i did $$s_k - s_{k-1}$$ and got 6k -3, which is the answer given in the text.

my question is, shouldnt $$s_k = u_k$$ so we substitute n = 1 in 3n^2 - 1, we get 2. but if we put k = 1 in $$u_k$$ we get 3..

where did i go wrong

• And, for $k>1$, we have $u_k=6(k-1)+3=6k-3$, but not for $k=1$. – Dietrich Burde Apr 20 at 14:11
• how do we know k > 1?/ – Vanessa Apr 20 at 14:17
• $s_{k-1}$ only makes sense for $k>1$. – Dietrich Burde Apr 20 at 14:18
• oh ya, thank youuu – Vanessa Apr 20 at 14:30
• The upper limit in the first sum should be $n$, not $\infty$. It would be good to combine that with the next equation to say $s_n=$ the sum = $3n^2-1$ – Ross Millikan Apr 20 at 14:41

It might be helpful to realize, that $$u_n$$ looks as follows:

$$u_n = \begin{cases} 2 & n= 1 \\ 6n-3 & n> 2 \end{cases}$$

The "suprising" part in this exercise is, that the first term $$u_1=s_1$$ does not follow the general rule $$6k-3$$ for the other terms of the sequence.

Summing the $$u_n$$ serves to verify the result:

$$s_n = \sum_{k=1}^n u_k = 2+ \sum_{k=2}^n(6k-3)= 2-3(n-1)+6\underbrace{\sum_{k=2}^n k}_{=\frac{n(n+1)}{2}-1}=3n^2-1$$

Taking the difference of successive $$s$$'s is the way to isolate one term as you have done. You can say $$s_n=\sum_{k=1}^nu_k$$ so $$s_n-s_{n-1}=u_n$$ because $$s_{n-1}$$ has all the terms up to $$u_{n-1}$$. All the terms except $$u_n$$ cancel in the subtraction. Then $$(3n^2-1)-(3(n-1)^2-1)=6n-3$$

If you substitute a value for $$n$$ you can find whatever term you want.