The coefficient of $x^{30}$ in the expansion of $(1 + 2 x + 3 x^{2} + . . . . 21 x^{20})^{2}$ is

2706

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2450

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1481

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256

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Solution

The correct option is C $2706$
$S = (1 + 2 x + 3 x^{2} + . . . . .21 x^{20}) (1 + 2 x + 3 x^{2} + . . . . . .21 x^{20})$
We can get $x^{30}$ by getting a pair of coefficient of $x$ such that they collectively form $x^{30} .$
For eg. $(21 x^{20} \times 11 x^{10}), (20 x^{19} \times 12 x^{11})$ and so on.
Hence, the coefficient,
$= (21 \times 11) + (20 \times 12) + (19 \times 13) + . . . . . . (11 \times 21)$
$= \sum_{r = 0}^{10} (21 - r) (11 + r) = \sum_{r = 0}^{10} (231 + 10 r - r^{2})$
$= \sum_{r = 0}^{10} 231 + 10 \sum_{r = 0}^{10} 0 r - \sum_{r = 0}^{10} r^{2}$
$= 231 \times 11 + 10 \times {\frac{(10) \times (10 + 1)}{2}} - \frac{10 \times (10 + 1) \times (20 + 1)}{6}$
$= 2541 + 550 - 385$
$= 2706.$
Hence, the answer is $2706.$

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Similar questions

x^{30}

(1 + 2 x + 3 x^{2} + . . . . .21 x^{20})^{2}

x^{5}

(1 + x)^{21} + (1 + x)^{22} + . . . . . + (1 + x)^{30}

The coefficient of x^5 in the expansion of $(1 + x)^{21} + (1 + x)^{22} + . . . . . + (1 + x)^{30}$ is

x^{4}

\frac{(1 - 3 x)^{2}}{(1 - 2 x)}

x^{5}

(1 + x)^{21} + (1 + x)^{22} + \dots \dots . . + (1 + x)^{30}

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