Show that $(1c_1 + c_2) + (2c_1 + c_2) + ... + (nc_1 + c_2)=\frac{c_1n(n+1)}{2} +nc_2$

Question

How do you convert / factor the following sum:

$(1c_1 + c_2) + (2c_1 + c_2) + ... + (nc_1 + c_2)$

Into:

$\dfrac{c_1n(n+1)}{2} +nc_2$

Answer 1

It is time to use important properties: associativity (and commutativity) of the addition, and the distributivity of products over adds. One calls that "linearity".

Your sum is thus $c_1$ times $(1+2+3+\cdots+n)$ plus $c_2$, $n$ times. For the intuition behind $(1+2+3+\cdots+n)$ , I am fond of visual proofs:

The sum of the first $n$ integers is $n(n+1)/2$ (half the size of the rectangle on the right). Hence, the result is:

$$ c_1\frac{n(n+1)}{2}+c_2 n$$

Answer 2

Hint : $$\sum_{r=1}^n r =\frac{n(n+1)}{2}$$

Answer 3

rewrite the sum in the form $$c_1(1+2+...+n)+c_2(1+1+...+1)$$ using that $$\sum_{i=1}^ni=\frac{n(n+1)}{2}$$ and $$1+1+...+1=n\cdot 1$$ you will get the desired result.

Answer 4

If you rearrange the terms of the sum, by gathering all the $c_1$'s and all the $c_2$, one of the terms of your result appear immediately, and the other is one of the most well known sums.

Answer 5

\begin{align*} \sum_{k=1}^{n} (k c_{1}+c_{2}) &= c_{1} \left( \sum_{k=1}^{n} k \right)+ c_{2} \left( \sum_{k=1}^{n} 1 \right) \\ &= \frac{n(n+1)}{2} c_{1}+n c_{2} \end{align*}