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Mathematics Question on Sequence and series

If the sum of the first ten terms of the series (135)2+(225)2+(315)2+42+(445)2+....,\left(1 \frac{3}{5}\right)^{2} + \left(2 \frac{2}{5}\right)^{2} + \left(3 \frac{1}{5}\right)^{2} + 4^{2} + \left(4 \frac{4}{5}\right)^{2} + .... , is 165m,\frac{16}{5} m , then mm is equal to :

A

102

B

101

C

100

D

99

Answer

101

Explanation

Solution

(135)2+(225)2+(315)2+42+(445)2+.....\left(1 \frac{3}{5}\right)^{2} + \left(2 \frac{2}{5}\right)^{2} + \left(3 \frac{1}{5}\right)^{2} + 4^{2} + \left(4 \frac{4}{5}\right)^{2} + ..... upto 1010 terms
=(85)2+(125)2+(165)2+(205)2+(245)2+...= \left(\frac{8}{5}\right)^{2} + \left(\frac{12}{5}\right)^{2} + \left(\frac{16}{5}\right)^{2} + \left(\frac{20}{5}\right)^{2} + \left(\frac{24}{5}\right)^{2} + ... upto 1010 terms.
(8)2+(12)2+(16)2+...\left(8\right)^{2} + \left(12\right)^{2} + \left(16\right)^{2} + ... up to 1010 terms
Tn[4(n+1)]2T_{n} \left[4\left(n+1\right)\right]^{2} where n varies from 11 to 1010.
=16(n2+2n+1)= 16\left(n^{2} + 2n + 1\right)
∑Tn=∑n=11016(n2+2n+1)\displaystyle\sum T_{n} = \sum^{10}_{n=1} 16 \left(n^{2} + 2n + 1\right)
=16[385+55(2)+10]= 16 \left[385 + 55\left(2\right) + 10\right]
=16(505)= 16\left(505\right)
or
∑n=110n2=n(n+1)(2n+1)6=10×11×216=385\displaystyle\sum^{10}_{n=1} n^{2} = \frac{n\left(n+1\right)\left(2n+1\right)}{6} = \frac{10 \times11 \times21}{6} = 385
∑n=110n=n(n+1)2=10×112=55\displaystyle\sum^{10}_{n=1} n = \frac{n\left(n+1\right)}{2} = \frac{10 \times11}{2} = 55
∑n=1101=n=10\displaystyle \sum^{10}_{n=1} 1 = n = 10
∴(85)2+(125)2+(165)2+.....\therefore \left(\frac{8}{5}\right)^{2} + \left(\frac{12}{5}\right)^{2} + \left(\frac{16}{5}\right)^{2} + ..... upto 1010 terms =16×50525= \frac{16 \times505}{25 }
It is given that 16×50525=165 m\frac{16 \times505}{25} = \frac{16}{5}\, m
∴m=5055=101\therefore m = \frac{505}{5} = 101