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JEE Main · 2025 · 03 Apr 2025 · Shift 2 · Binomial Theorem

Let

(1 + x + x^{2})^{10} = a_{0} + a_{1} x + a_{2} x^{2} + \dots . + a_{20} x^{20}

. If

(a_{1} + a_{3} + a_{5} + \dots . + a_{19}) - 11 a_{2} = 121 k

, then k is equal to ______ .

Answer: 239

Solution

Putting

x = 1

gives

a_{0} + a_{1} + \dots + a_{20} = 3^{10}

and

x = - 1

gives the alternating sum equal to

1

; subtracting,

a_{1} + a_{3} + \dots + a_{19} = \frac{3 ^{10} - 1}{2} = 29524

. The coefficient of

x^{2}

in

(1 + x + x^{2})^{10}

is

a_{2} = (1 10) + (2 10) = 10 + 45 = 55

. Then

\frac{29524 - 11 \cdot 55}{121} = \frac{28919}{121}

, so

k = 239

.

JEE Main · 2025 · 03 Apr 2025 · Shift 2 · Integer answer

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