Computing sums and double sums I would like to check if my answer is correct:
1) Compute $$\sum_{i=1}^n(3i+4)$$. I got $(3n^2 + 11n) / 2$ using the property that $\sum_{i=1}^n i = n(n+1)/2$
2) I'm unsure how to approach double summations like this: $$\sum_{x = 1}^n\sum_{y=1}^n(x+y−1)$$. Please help explain and the correct answer would be appreciated!
 A: 1) Looks fine to me.
2) Notice this:
$$\begin{array}{}\displaystyle\sum_{x=1}^n\sum_{y=1}^n(x+y-1)&\displaystyle\vphantom{\cfrac11}=\sum_{x=1}^n\sum_{y=1}^nx&\displaystyle+\sum_{x=1}^n\sum_{y=1}^ny&\displaystyle-\sum_{x=1}^n\sum_{y=1}^n1\\\vphantom{\cfrac11}&\displaystyle=\sum_{x=1}^nx\sum_{y=1}^n1&\displaystyle+\sum_{x=1}^n1\sum_{y=1}^ny&\displaystyle-\sum_{x=1}^n1\sum_{y=1}^n1\\\vphantom{\cfrac11}&\displaystyle=\frac{n(n+1)}2\cdot n&\displaystyle+n\cdot\frac{n(n+1)}2&\displaystyle-n\cdot n\\&=n^3\end{array}$$
A: 1) You got that right.
2) Do the sum for $y=1$ to $n$ as if $x$ is a constant. Then sum the result over the values of $x$
$\sum_{y=1}^n(x+y−1)= \dfrac {n(n+1)}{2}+n(x-1)= \dfrac {n(n+1)-2n}{2}+nx=\dfrac {n(n-1)}{2}+nx$
$\sum_{x = 1}^n\sum_{y=1}^n(x+y−1)=\dfrac {n^2(n-1)}{2}+n \dfrac {n(n+1)}{2}=n^3$
A: A few tricks: 


*

*the sum of the terms is always the product of the number of terms and the average value of the terms,

*the average value of a constant term is that constant,

*the average value of a linear term is the mean of the extreme values.
Then by 1. and 3.,
$$\sum_{i=1}^n(3i+4)=n\frac{3+4+3n+4}2.$$
For the other sum, you can decompose in three double sums, on $x$, $y$ and $-1$.
Then by 1., 2. and 3.,
$$\sum_{x = 1}^n\sum_{y=1}^n(x+y−1)=2n\frac{n^2+n}2-n^2.$$

Justification:
In the double sum of $x$, the $x$ is a constant for the inner sum, which then equals $nx$. The outer sum yields 
$$n\frac{n^2+n}2.$$
By symmetry, the sum on $y$ equals that on $x$, hence the factor $2$.
