Question

Find the sum of the first $40$ terms of the series $1^2-2^2+3^2-4^2+.....$

Answer 1

The given arithmetic series is $1^2-2^2+3^2-4^2+......40$ terms which can be rewritten as $(1-4)+(9-16)+....20$ terms. Thus the arithmetic series becomes

$(-3)+(-7)+......20$ terms where the first term is $a_1=-3$, second term is $a_2=-7$ and the number of terms $n=20$.

We find the common difference $d$ by subtracting the first term from the second term as shown below:

$d=a_2-a_1=-7-(-3)=-7+3=-4$

We know that the sum of an arithmetic series with first term $a$ and common difference $d$ is $S_n=\frac{n}{2}[2a+(n-1\left)d\right]$

Now to find the sum of series, substitute $n=20,a=-3$ and $d=-4$ in $S_n=\frac{n}{2}[2a+(n-1\left)d\right]$ as follows:

$S_{20}=\frac{20}{2}\left[\right(2\times -3)+(20-1\left)\right(-4\left)\right]=10[-6+(19\times -4\left)\right]=10(-6-76)=10\times -82=-820$

Hence, the sum of the given arithmetic series is $-820$.