If the sum of the first ten terms of the series (1 (3/5))2 + (2 (2/5))2 + (3 (1/5))2 + 42 + (4 (4/5))2 + .... , is (16/5) m , then m is equal to :

Question

If the sum of the first ten terms of the series (1 (3/5))2 + (2 (2/5))2 + (3 (1/5))2 + 42 + (4 (4/5))2 + .... , is (16/5) m , then m is equal to : (A) 102 (B) 101 (C) 100 (D) 99

Tardigrade Admin · Accepted Answer

(1 (3/5))2 + (2 (2/5))2 + (3 (1/5))2 + 42 + (4 (4/5))2 + ..... upto 10 terms = ((8/5))2 + ((12/5))2 + ((16/5))2 + ((20/5))2 + ((24/5))2 + ... upto 10 terms. (8)2 + (12)2 + (16)2 + ... up to 10 terms Tn [4(n+1)]2 where n varies from 1 to 10. = 16(n2 + 2n + 1) displaystyle∑ Tn = ∑10n=1 16 (n2 + 2n + 1) = 16 [385 + 55(2) + 10] = 16(505) or displaystyle∑10n=1 n2 = (n(n+1)(2n+1)/6) = (10 ×11 ×21/6) = 385 displaystyle∑10n=1 n = (n(n+1)/2) = (10 ×11/2) = 55 displaystyle ∑10n=1 1 = n = 10 ∴ ((8/5))2 + ((12/5))2 + ((16/5))2 + ..... upto 10 terms = (16 ×505/25 ) It is given that (16 ×505/25) = (16/5) m ∴ m = (505/5) = 101

Q. If the sum of the first ten terms of the series (153​)2+(252​)2+(351​)2+42+(454​)2+...., is 516​m, then m is equal to :

102

101

100

99

Q. 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 :