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Question Number 114638 by bobhans last updated on 20/Sep/20

$$Givena=\frac{1^2+2^2+3^2+...+{16}^2-16}{1.3+2.4+3.5+...+15.17}$$$$c=(1-\frac{1}{2}).(1-\frac{1}{3}).(1-\frac{1}{4}).(1-\frac{1}{5}).$$$$(1+\frac{1}{5})(1+\frac{1}{4})(1+\frac{1}{3})(1+\frac{1}{2}).$$$$finda\times c=$$

Answered by bemath last updated on 20/Sep/20

$$\left(1\right)c=\frac{1}{2}\times \frac{2}{3}\times \frac{3}{4}\times \frac{4}{5}\times \frac{6}{5}\times \frac{5}{4}\times \frac{4}{3}\times \frac{3}{2}$$$$c=\frac{1}{5}\times \frac{6}{2}=\frac{3}{5}$$$$a=\frac{\underset{k=1}{\sum ^{16}}{k}^2-16}{\underset{k=1}{\sum ^{15}}k(k+2)}=\frac{\frac{16.17.33}{6}-16}{\frac{15.16.31}{6}+2\left(\frac{15.16}{2}\right)}$$$$=\frac{8.17.11-16}{5.8.31+15.16}=\frac{8(17.11-2)}{5.8(31+6)}$$$$=\frac{185}{5.37}=1$$$$thena\times c=1\times \frac{3}{5}=\frac{3}{5}$$

Commented by PRITHWISH SEN 2 last updated on 20/Sep/20

$$\mathrm{c}=\frac{3}{5}$$

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