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JEE Main · 2023 · 24 Jan 2023 · Shift 1 · Sequences and Series

If

\frac{1 ^{3} + 2 ^{3} + 3 ^{3} + \dots upto n terms}{1 \cdot 3 + 2 \cdot 5 + 3 \cdot 7 + \dots upto n terms} = \frac{9}{5}

, then the value of

n

is

Answer: 5

Solution

The numerator is

\sum_{r = 1}^{n} r^{3} = (\frac{n ( n + 1 )}{2})^{2}

. The denominator has

r

-th term

r (2 r + 1) = 2 r^{2} + r

, so it equals

\frac{2 n ( n + 1 ) ( 2 n + 1 )}{6} + \frac{n ( n + 1 )}{2} = \frac{n ( n + 1 ) ( 4 n + 5 )}{6}

. The ratio simplifies to

\frac{3 n ( n + 1 )}{2 ( 4 n + 5 )} = \frac{9}{5}

, giving

15 n (n + 1) = 18 (4 n + 5)

, i.e.

5 n^{2} - 19 n - 30 = 0

, which factors as

(n - 5) (5 n + 6) = 0

. The admissible value is

n = 5

.

JEE Main · 2023 · 24 Jan 2023 · Shift 1 · Integer answer

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