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JEE Main · 2024 · 04 Apr 2024 · Shift 2 · Sequences and Series

The value of

\frac{1 \times 2 ^{2} + 2 \times 3 ^{2} + \dots + 100 \times ( 101 ) ^{2}}{1 ^{2} \times 2 + 2 ^{2} \times 3 + \dots + 10 0 ^{2} \times 101}

is

(A) $\frac{305}{301}$ ✓ Correct
(B) $\frac{306}{305}$
(C) $\frac{31}{30}$
(D) $\frac{32}{30}$

Answer: (A)

Solution

k = 1 \sum 100 k (k + 1)^{2} = \sum (k^{3} + 2 k^{2} + k) = (\frac{n ( n + 1 )}{2})^{2} + \frac{n ( n + 1 ) ( 2 n + 1 )}{3} + \frac{n ( n + 1 )}{2}_{n = 100} = 25502500 + 676700 + 5050 = 26184250

k = 1 \sum 100 k^{2} (k + 1) = \sum (k^{3} + k^{2}) = (\frac{n ( n + 1 )}{2})^{2} + \frac{n ( n + 1 ) ( 2 n + 1 )}{6}_{n = 100} = 25502500 + 338350 = 25840850

Ratio

= \frac{26184250}{25840850} = \frac{305}{301}

(dividing through by 101×850).

JEE Main · 2024 · 04 Apr 2024 · Shift 2 · Single correct option

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