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JEE Main · 2024 · 06 Apr 2024 · Shift 1 · Binomial Theorem

If the second, third and fourth terms in the expansion of

(x + y)^{n}

are 135, 30 and

\frac{10}{3}

, respectively, then

6 (n^{3} + x^{2} + y)

is equal to ____.

Answer: 806

Solution

With

T_{2} = 135, T_{3} = 30, T_{4} = \frac{10}{3}

:

\frac{^{n} C _{1} ^{n} C _{3}}{( ^{n} C _{2} ) ^{2}} = \frac{T _{2} / T _{3}}{T _{3} / T _{4}} = \frac{9/2}{9} = \frac{1}{2}

, which gives

n = 5

.
Then

\frac{x}{y} = 9

and

5 x^{4} y = 135 \Rightarrow x^{4} y = 27

; with

x = 9 y

this gives

y = \frac{1}{3}, x = 3

.

6 (n^{3} + x^{2} + y) = 6 (125 + 9 + \frac{1}{3}) = 6 \cdot \frac{403}{3} = 806.

JEE Main · 2024 · 06 Apr 2024 · Shift 1 · Integer answer

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