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Question Number 114433 by bemath last updated on 19/Sep/20

$$findsumoftheseries$$$$1^2-3^2+5^2-7^2+9^2-{11}^2+...+{(4n-3)}^2-{(4n-1)}^2$$

Answered by 1549442205PVT last updated on 19/Sep/20

$$\mathrm{S}=1+(5^2-3^2)+(9^2-7^2)+({13}^2-{11}^2)+...[{(4\mathrm{n}-3)}^2-{(4\mathrm{n}-5)}^2]-{(4\mathrm{n}-1)}^2$$$$=1+(5+3)(5-3)+(9+7)(9-7)+(13+11)(13-11)+...+(8\mathrm{n}-8).2-{(4\mathrm{n}-1)}^2$$$$=1+2[8+16+24+...+(8\mathrm{n}-8)]-{(4\mathrm{n}-1)}^2$$$$=1+2\left[\frac{(8+8\mathrm{n}-8)(\mathrm{n}-1)}{2}\right]-{(4\mathrm{n}-1)}^2$$$$=1+8\mathrm{n}(\mathrm{n}-1)-{16\mathrm{n}}^2+8\mathrm{n}-1$$$$\mathrm{S}=-{8\mathrm{n}}^2$$

Commented by bemath last updated on 19/Sep/20

$$yes..gavekudos✓$$

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