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