The sum of n terms of $(2n-1)+2(2n-3)+3(2n-5)+...$ is

Question

We can split the nth term into

{nth term of 1,2,3,4...}{nth term of 2n-1,2n-3,2n-5...}

nth term of 2n-1,2n-3... $$a_n = 2n-1 + (n-1)(-2)$$ Since common difference is -2

$$a_n = 1$$

There nth term of the series is $$T_n=n$$ So $$S_n=\sum_{n=1}^n = \frac{(n)(n+1)}{2}$$

This answer is wrong, but I needed to show what I thought was right. How should I solve it correctly?

Answer 1

Observe that

$$2S=\sum_{r=1}^n(2r)(2n+1-2r)=2(2n+1)\sum_{r=1}^nr-4\sum_{...}r^2$$

Use The sum of squares of the first $n$ natural numbers.

Answer 2

The $k$th term in the sum is given by the product of the $k$th term in the seuqnence $(1,2,3,4,\dots)$ with the $k$th term of the sequence $(2n-1,2n-3,2n-5,\dots)$. Note here that $n$ and $k$ are different. $n$ is a fixed constant, $k$ is not and indicates the position of the term we are looking at. The first factor is obviously $k$. To see the second factor, note that the $k$th term of the sequence $(1,3,5,7,\dots)$ is given by $2k-1$, so the $k$th term of $(2n-1,2n-3,2n-5,\dots)$ is simply given by $2n-(2k-1)=2n-2k+1$. Hence, the $k$th term in the sum is just $k(2n-2k+1)$.

The sum you're dealing with is $$\sum_{k=1}^nk(2n-2k+1)=(2n+1)\sum_{k=1}^nk-2\sum_{k=1}^nk^2.$$ Both sums $\sum k$ and $\sum k^2$ are well known (for the latter, just search for "sum of first $n$ squares" if you don't already know the formula). The rest should be easy.

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Note, your solution is wrong because the $k$th term in the sequence is not in fact $(2k-1)+(k-1)(-2)$. It should be $k(2n-2k+1)$. I suspect that the main problem that you have is that you are confusing the indexing variable $k$ (indicating which term in the sequence you are looking at) and the constant $n$. We are not looking at the $n$th term of the sequence, but the $k$th term. This misconception is quite obvious when you wrote $\sum_{n=1}^n$ in the last line too. Can you try to convince yourself why $k$ and $n$ must be treated differently?