Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 154791 by azadsir last updated on 21/Sep/21

Answered by Ar Brandon last updated on 24/Sep/21

$$S=9^2+7^2+5^2+3^2+1^2+{(-1)}^2+{(-3)}^2+\cdot \cdot \cdot +{(-25)}^2$$$$=9^2+7^2+5^2+3^2+1^2+1^2+3^2+\cdot \cdot \cdot +{25}^2$$$$=\underset{n=1}{\sum ^5}{(2n-1)}^2+\underset{n=1}{\sum ^{13}}{(2n-1)}^2$$$$=\underset{n=1}{\sum ^5}(4n^2-4n+1)+\underset{n=1}{\sum ^{13}}(4n^2-4n+1)$$$$=\frac{2\left(5\times 6\times 11\right)}{3}-2\left(5\times 6\right)+5$$$$+\frac{2\left(13\times 14\times 27\right)}{3}-2\left(13\times 14\right)+13$$$$=220-60+5+3276-364+13$$$$=3441-351=3090$$

Commented by Ar Brandon last updated on 24/Sep/21

$${\mathrm{You}}^{\prime }\mathrm{re}\mathrm{right}.{\mathrm{It}}^{\prime }\mathrm{s}\mathrm{supposed}\mathrm{to}\mathrm{be}$$$$\underset{n=1}{\sum ^5}\mathrm{instead}\mathrm{of}\underset{n=1}{\sum ^4}$$$$3009+9^2=3090$$

Commented by SLVR last updated on 24/Sep/21

$$Excuse..me..errorincaliculation$$$$itis3090corret...insteadof3009$$

Commented by SLVR last updated on 24/Sep/21

$$sir....donotmistakeme...$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com