Sequence-And-Series Question 665
Question: If the sum of the 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:
Options:
A) 100
B) 99
C) 102
D) 101
Show Answer
Answer:
Correct Answer: D
Solution:
[d] $ {{( \frac{8}{5} )}^{2}}+{{( \frac{12}{5} )}^{2}}+{{( \frac{16}{5} )}^{2}}+{{( \frac{20}{5} )}^{2}}….+{{( \frac{44}{5} )}^{2}} $ $ S=\frac{16}{25}( 2^{2}+3^{2}+4^{2}+….+11^{2} ) $ $ =\frac{16}{25}( \frac{11(11+1)(22+1)}{6}-1 ) $ $ =\frac{16}{25}\times 505=\frac{16}{5}\times 101 $
$ \Rightarrow \frac{16}{5}m=\frac{16}{5}\times 101\Rightarrow m=101. $
BETA
