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JEE Main · 2019 · 10 Apr 2019 · Shift 1 · Sequences and Series

The sum

\frac{3 \times 1}{1 ^{2}} + \frac{5 \times ( 1 ^{3} + 2 ^{3} )}{1 ^{2} + 2 ^{2}} + \frac{7 \times ( 1 ^{3} + 2 ^{3} + 3 ^{3} )}{1 ^{2} + 2 ^{2} + 3 ^{2}} + \dots\dots .

Upto

1 0^{th}

term, is

(A)620
(B)660✓ Correct
(C)680
(D)600

Answer: (B)

Solution

The

n

th term is

T_{n} = \frac{( 2 n + 1 ) \sum k ^{3}}{\sum k ^{2}}

summed to

n

. Using

\sum k^{3} = (\frac{n ( n + 1 )}{2})^{2}

and

\sum k^{2} = \frac{n ( n + 1 ) ( 2 n + 1 )}{6}

, it simplifies to

T_{n} = \frac{( 2 n + 1 ) \frac{n ^{2} ( n + 1 ) ^{2}}{4}}{\frac{n ( n + 1 ) ( 2 n + 1 )}{6}} = \frac{3 n ( n + 1 )}{2}

. Summing the first ten terms,

S_{10} = \frac{3}{2} \sum_{n = 1}^{10} n (n + 1) = \frac{3}{2} \cdot \frac{10 \cdot 11 \cdot 12}{3} = 660

.

JEE Main · 2019 · 10 Apr 2019 · Shift 1 · Single correct option

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