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JEE Main · 2024 · 09 Apr 2024 · Shift 2 · Binomial Theorem

The sum of the coefficient of

x^{2/3}

and

x^{- 2/5}

in the binomial expansion of

(x^{2/3} + \frac{1}{2} x^{- 2/5})^{9}

is :

(A)21/4✓ Correct
(B)69/16
(C)63/16
(D)19/4

Answer: (A)

Solution

General term of

(x^{2/3} + \frac{1}{2} x^{- 2/5})^{9}

:

T_{r + 1} = (r 9) (x^{2/3})^{9 - r} (\frac{1}{2} x^{- 2/5})^{r} = (r 9) (\frac{1}{2})^{r} x^{\frac{2 ( 9 - r )}{3} - \frac{2 r}{5}}

Exponent

= 6 - \frac{16 r}{15}

.

For exponent

= \frac{2}{3}

:

6 - \frac{16 r}{15} = \frac{2}{3} ⟹ r = 5

; coefficient

= (5 9) (\frac{1}{2})^{5} = 126 \times \frac{1}{32} = \frac{63}{16}

.

For exponent

= - \frac{2}{5}

:

6 - \frac{16 r}{15} = - \frac{2}{5} ⟹ r = 6

; coefficient

= (6 9) (\frac{1}{2})^{6} = 84 \times \frac{1}{64} = \frac{21}{16}

.

Sum = \frac{63}{16} + \frac{21}{16} = \frac{84}{16} = \frac{21}{4}

JEE Main · 2024 · 09 Apr 2024 · Shift 2 · Single correct option

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