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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

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  • $\begingroup$ And, for $k>1$, we have $u_k=6(k-1)+3=6k-3$, but not for $k=1$. $\endgroup$
    – Dietrich Burde
    Commented Apr 20, 2019 at 14:11
  • $\begingroup$ how do we know k > 1?/ $\endgroup$
    – Vanessa
    Commented Apr 20, 2019 at 14:17
  • $\begingroup$ $s_{k-1}$ only makes sense for $k>1$. $\endgroup$
    – Dietrich Burde
    Commented Apr 20, 2019 at 14:18
  • $\begingroup$ oh ya, thank youuu $\endgroup$
    – Vanessa
    Commented Apr 20, 2019 at 14:30
  • $\begingroup$ 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 = $3n^2-1$ $\endgroup$
    – Ross Millikan
    Commented Apr 20, 2019 at 14:41

2 Answers 2

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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:

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

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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.

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