Summation notation for $1*3+2*4+3*5+...+40*42$

Question

For this pattern $1*3+2*4+3*5+...+40*42$

my summation notation was: $\sum_{i=1}^n i(i+2), n = 40$

Which gives me the answer of $23780.$

Is this correct notation?

Answer 1

The summation notation is right. The sum is also correct, since

$$\sum_{i=1}^ni(i+2)=\sum _{i=1}^n i^2+2i=\sum_{i=1}^ni^2+2\sum_{i=1}^ni=\frac{n(n+1)(2n+1)}{6}+2\frac{n(n+1)}{2}.$$ After you set $n=40$, you get $23780$.

Answer 2

Your notation appears to be correct, assuming that, as you did, the $i$th term of the sum is $i(i+2)$; to evaluate it, expand the brackets inside the sum then write the sum as a sum of known, standard sums.

Answer 3

That is right.

To get an expression for the sum with $n$ terms,

$\begin{array}\\ \sum_{i=1}^ni(i+2) &=\sum_{i=1}^n(i^2+2i)\\ &=\sum_{i=1}^n((i+1)^2-1)\\ &=\sum_{i=2}^{n+1}(i^2-1)\\ &=\sum_{i=2}^{n+1}i^2-\sum_{i=2}^{n+1}1\\ &=\sum_{i=1}^{n+1}i^2-1-n\\ &=\dfrac{(n+1)(n+2)(2n+3)}{6}-(n+1)\\ &=\dfrac{(n+1)((n+2)(2n+3)-6)}{6}\\ &=\dfrac{(n+1)(2n^2+7n)}{6}\\ &=\dfrac{n(n+1)(2n+7)}{6}\\ \end{array} $

If $n=40$, this is

$\begin{array}\\ \dfrac{n(n+1)(2n+7)}{6} &=\dfrac{40\ 41\ 87}{6}\\ &=20\ 41\ 29\\ &=20 (40+1)(30-1)\\ &=20 (1200-10-1)\\ &=20 (1189)\\ &= 23780\\ \end{array} $