Discussion Forum

If the sum of first ten terms of the series $(1\frac{3}{5})^{2}+(2\frac{2}{5})^{2}+(3\frac{1}{5})^{2}+4^{2}+(4\frac{4}{5})^{2}+ ... is \frac{16}{5}m,$, then m is equal to

Answer:

Explanation:

Let S₁₀ be the sum of the first ten terms of the series, then we have 

$S_{10}= (1\frac{3}{5})^{2}+(2\frac{2}{5})^{2}+(3\frac{1}{5})^{2}+4^{2}+(4\frac{4}{5})^{2}+... to$ 10 terms

  $= (\frac{8}{5})^{2}+(\frac{12}{5})^{2}+(\frac{16}{5})^{2}+4^{2}+(\frac{24}{5})^{2}+... to$ 10 terms

= $\frac{1}{5^{2}}({8}^{2}+{12}^{2}+16^{2}+{20}^{2}+{24}^{2}... to$  10 terms

=$\frac{4^{2}}{5^{2}}({2}^{2}+{3}^{2}+4^{2}+{5}^{2}+... to$  to  10 terms)

= $\frac{4^{2}}{5^{2}}({2}^{2}+{3}^{2}+4^{2}+{5}^{2}+... +11^{2})$

= $\frac{16}{25}(({1}^{2}+{2}^{2}+3^{2}+{4}^{2}+... +11^{2})-1^{2})$

=$\frac{16}{25}(\frac{11.(11+1)(2.11+1)}{6}-1)$

 =$\frac{16}{25}(506-1)=\frac{16}{25}\times 505$

$\Rightarrow \frac{16}{5}m=\frac{16}{25}\times 505=101$