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JEE Main · 2023 · 25 Jan 2023 · Shift 1 · Binomial Theorem

If

a_{r}

is the coefficient of

x^{10 - r}

in the Binomial expansion of

(1 + x)^{10}

, then

\sum_{r = 1}^{10} r^{3} (\frac{a _{r}}{a _{r - 1}})^{2}

is equal to

(A)4895
(B)1210✓ Correct
(C)5445
(D)3025

Answer: (B)

Solution

Since

a_{r} = (10 - r 10) = (r 10)

,

\frac{a _{r}}{a _{r - 1}} = \frac{11 - r}{r}

, so

\sum_{r = 1}^{10} r^{3} (\frac{11 - r}{r})^{2} = \sum_{r = 1}^{10} r (11 - r)^{2} = \sum_{r = 1}^{10} (121 r - 22 r^{2} + r^{3})

. Using

\sum r = 55

,

\sum r^{2} = 385

and

\sum r^{3} = 3025

, this equals

6655 - 8470 + 3025 = 1210

.

JEE Main · 2023 · 25 Jan 2023 · Shift 1 · Single correct option

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