Question

If the sum of the first $11$ terms of the series ${\left(1\frac{4}{7}\right)}^2+{\left(1\frac{5}{7}\right)}^2+{\left(1\frac{6}{7}\right)}^2+2^2+\left(2\frac{1}{7}\right)+.........$ $\frac{11}{7}\lambda$ then ,$\lambda$ is equal to

• A. $36$
• B. $37$
• C. $38$
• D. $39$

Answer 1

The correct option is C $38$

${\left(1\frac{4}{7}\right)}^2+{\left(1\frac{5}{7}\right)}^2+{\left(1\frac{6}{7}\right)}^2+2^2+{\left(2\frac{1}{7}\right)}^2+{\left(2\frac{2}{7}\right)}^2+{\left(2\frac{3}{7}\right)}^2+....3^2={\left(\frac{11}{7}\right)}^2+{\left(\frac{12}{7}\right)}^2+{\left(\frac{13}{7}\right)}^2+{\left(\frac{14}{7}\right)}^2+.....{\left(\frac{21}{7}\right)}^2{S}_n=\frac{n(n+1)(2n+1)}{6}(1^2+2^2+3^2+....n^2)Totalsum=\frac{1}{49}(S_{21}-S_{10})=\frac{1}{49}(\frac{21(21+1)(42+1)}{6}-\frac{10(10+1)(20+1)}{6})=\frac{1}{49}(\frac{21\times 22\times 43}{6}-\frac{10\times 11\times 21}{6})=\frac{1}{49}(7\times 11\times 43-5\times 11\times 7)=\frac{1}{49}\left\{77\right(43-5\left)\right\}=\frac{1}{49}\left(77\times 38\right)=\frac{11}{7}\left(38\right)\Rightarrow \lambda =38$