Question and Answers Forum

All Questions      Topic List

Arithmetic Questions

Previous in All Question      Next in All Question      

Previous in Arithmetic      Next in Arithmetic      

Question Number 17767 by Joel577 last updated on 10/Jul/17

$$\underset{n=1}{\sum ^{30}}(n^2+1)=$$$$\left(\mathrm{A}\right)\underset{n=1}{\sum ^{15}}(2n^2+30n+224)$$$$\left(\mathrm{B}\right)\underset{n=1}{\sum ^{15}}(2n^2+30n+225)$$$$\left(\mathrm{C}\right)\underset{n=1}{\sum ^{15}}(2n^2+30n+226)$$$$\left(\mathrm{D}\right)\underset{n=1}{\sum ^{15}}(2n^2+30n+227)$$$$\left(\mathrm{E}\right)\underset{n=1}{\sum ^{15}}(2n^2+30n+228)$$

Answered by Tinkutara last updated on 11/Jul/17

$$\underset{n=1}{\sum ^{30}}(n^2+1)=1^2+1+2^2+1+...+{30}^2+1$$$$=\frac{30\times 31\times 61}{6}+30=9485$$$$\mathrm{Now}\mathrm{option}\left(\mathrm{D}\right)=\frac{n(n+1)(2n+1)}{3}$$$$+15n(n+1)+227n$$$$=\frac{15\times 16\times 31}{3}+15\times 15\times 16+227\times 15$$$$=9485$$$$\mathrm{Hence}\mathrm{answer}\mathrm{is}\left(\mathrm{D}\right).$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com