Question

Find each sum

$-1+3+7+....+(4n-5)$

Short answer

The solution is$S_n=\frac{n}{2}(-6+4n)$

Step-by-step solution

Given information

$-1+3+7+....+(4n-5)$

Calculation

We have, $-1+3+7+....+(4n-5)$

So, $a_1=-1,d=4,a_n=4n-5$

Now use formula,

$$\begin{gathered} S_n=\frac{n}{2}(a_1+a_n) \\ =\frac{n}{2}(-1+4n-5) \\ =\frac{n}{2}(-6+4n) \end{gathered}$$

The sum is$S_n=\frac{n}{2}(-6+4n)$