Question

Find the sum of $1^2-2^2+3^2-4^2+5^2-6^2.......$?

Answer 1

Step 1: Calculate the sum of the series, if the number of terms is even

Given series: $1^2-2^2+3^2-4^2+5^2-6^2+.......$

Let the number of terms of the series be $n$, which is an even number.

Thus, $\left(1^2-2^2\right)+\left(3^2-4^2\right)+\left(5^2-6^2\right).......+\left[{\left(n-1\right)}^2-n^2\right]$

$$\begin{gathered} =\left(1-2\right)\left(1+2\right)+\left(3-4\right)\left(3+4\right)+\left(5-6\right)\left(5+6\right)+.......+\left[\left(n-1\right)-n\right]\left[\left(n-1\right)+n\right] \\ =\left(-1\right)\left(1+2\right)+\left(-1\right)\left(3+4\right)+\left(-1\right)\left(5+6\right)+......+\left(-1\right)\left[\left(n-1\right)+n\right] \\ =\left(-1\right)\left(1+2+3+4+5+6+.....+n\right) \end{gathered}$$

Use the formula of the sum of the first $n$ natural numbers.

$$\begin{gathered} =\left(-1\right)\frac{n\left(n+1\right)}{2} \\ =-\left(\frac{n^2+n}{2}\right) \end{gathered}$$

Hence, the sum of the series, when the number of terms is even, is $\mathrm{-}\mathrm{\left(}\frac{{\mathrm{n}}^{\mathrm{2}}\mathrm{+}\mathrm{n}}{\mathrm{2}}\mathrm{\right)}$.

Step 2: Calculate the sum of the series, if the number of terms is odd:

Given series: $1^2-2^2+3^2-4^2+5^2-6^2.......$

Let the number of terms of the series be $n$, which is an odd number.

Thus, $\left(1^2-2^2\right)+\left(3^2-4^2\right)+\left(5^2-6^2\right).......+\left[{\left(n-2\right)}^2-{\left(n-1\right)}^2\right]+n^2$

$$\begin{gathered} =\left(1-2\right)\left(1+2\right)+\left(3-4\right)\left(3+4\right)+\left(5-6\right)\left(5+6\right)+.......+\left[\left(n-2\right)-\left(n-1\right)\right]\left[\left(n-2\right)+\left(n-1\right)\right]+n^2 \\ =\left(-1\right)\left(1+2\right)+\left(-1\right)\left(3+4\right)+\left(-1\right)\left(5+6\right)+......+\left(-1\right)\left[\left(n-2\right)+\left(n-1\right)\right]+n^2 \\ =\left(-1\right)\left[1+2+3+4+5+6+.....+\left(n-1\right)\right]+n^2 \end{gathered}$$

Step 3: Calculate the sum of the first $n$ natural number

Use the formula of the sum of the first $n$ natural numbers.

$$\begin{gathered} =\left(-1\right)\frac{\left(n-1\right)\left[\left(n-1\right)+1\right]}{2}+n^2 \\ =-\frac{\left(n-1\right)n}{2}+n^2 \\ =-\left(\frac{n^2-n}{2}\right)+n^2 \\ =\frac{-n^2+n+2n^2}{2} \\ =\frac{n^2+n}{2} \end{gathered}$$

Hence, the sum of the series, when the number of terms is odd, is $\frac{n^2+n}{2}$.

Thus, in general, the sum of the series can be calculated by ${\left(-1\right)}^{n+1}\left(\frac{n^2+n}{2}\right)$

Therefore, the sum of the given series is ${\left(-1\right)}^{n+1}\left(\frac{n^2+n}{2}\right)$, where $n$ is an integer.