The sum of n terms of the series $1^2-2^2+3^2-4^2+5^2-6^2+....$ is

$$\begin{gathered} 1^2-2^2+3^2-4^2+5^2-6^2 ......... \mathrm{nterms}. \\ \mathrm{n} \mathrm{is} \mathrm{even} 1^2-2^2+3^2-4^2 ..... + {(\mathrm{n}-1)}^2-{\mathrm{n}}^2 \\ =-(1+2+3+4 .... + \mathrm{n}-1+\mathrm{n})=-\frac{\left(\mathrm{n}\right) (\mathrm{n}+1)}{2} \\ \mathrm{n} \mathrm{is} \mathrm{odd} 1^2-2^2+3^2-4^2 ..... + {(\mathrm{n}-2)}^2-{(\mathrm{n}-1)}^2+{\mathrm{n}}^2 \\ =-(1+2+3+4 ...... + \mathrm{n}-1)+{\mathrm{n}}^2 \\ = -\frac{\mathrm{n}(\mathrm{n}-1)}{2}+{\mathrm{n}}^2=\frac{2{\mathrm{n}}^2-{\mathrm{n}}^2+\mathrm{n}}{2}=\frac{\mathrm{n}(\mathrm{n}+1)}{2} \\ \therefore {(-1)}^{\mathrm{n}-1} \frac{\mathrm{n}(\mathrm{n}+1)}{2} \end{gathered}$$