Question

Match the series in column I with its value in column II

Column-I	Column-II
(A) The sum to 50 terms of the series $1+(1+2)+(1+2+3)+...$ is	(p) $9316$
(B) The sum to 10 terms of the series $(3^3–2^3)+(5^3–4^3)(1+2+3)+...$ is	(q) $4960$
(C) The sum to 16 terms of the series $1+(2+3)+(4+5+6)+......$ is	(r) $22100$
(D) The sum to 24 terms of the series $1^2+3^2+5^2+7^2+......$ is	(s) $18424$
	(t) Greater than $(1^3+2^3+......+{10}^3)$

• A. A(r, t), B(q, t), C(s, t), D(p, t)
• B. A(r, t), B(q, t), C(p, t), D(s, t)

Answer 1

The correct option is B A(r, t), B(q, t), C(p, t), D(s, t)

(A) $1+(1+2)+(1+2+3)+.....+\frac{n(n+1)}{2}$
$t_n=\frac{n^2+n}{2}=\frac{n(n+1)}{2}$
$S_n=\frac{n(n+1)(n+2)}{6}$

$S_{50}=\frac{50\times 51\times 52}{6}=22100$
(B) $(3^3–2^3)+(5^3–4^3)+....+\text{to}10$ terms
$t_n=(2n+1{)}^3–(2n{)}^3$
$=12n^2+6n+1$
$S_n=\frac{12n(n+1)(2n+1)}{6}+\frac{6n(n+1)}{2}+n$
$=2n(n+1)(2n+1)+3n(n+1)+n$
$S_{10}=2\times 10\times 11\times 21+3\times 10\times 11+10$
$=4620+330+10=4960$
(C) $1+(2+3)+(4+5+6).....\text{to}16$ terms
$=1+2+3+4+5+6+...\text{to}\frac{16\times 17}{2}$ terms
$=1+2+3+.......\text{to}136$ terms
$=\frac{136\times 137}{2}=9316$
(D) $1^2+3^2+5^2+7^2+......t0n\text{terms}=\frac{n(4n^2-1)}{3}$
set $n=24\text{to get}{S}_{24}=\frac{24\times (4\times {24}^2-1)}{3}$
$=8\times 2303=18424$