The co-efficient of $x^{401}$ in the expansions of $(1 + x + x^{2} + . . . . + x^{9})^{- 1}, (| x | < 1)$ is -

1

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Solution

The correct option is D $- 1$

${(1 + x + x^{2} + \dots x^{n})}^{- 1}$
$This is a GP with 10 terms$
$Sum of G P = a \frac{(1 - r^{n})}{(1 - r)}$
$= {[1 \frac{(1 - x^{n})}{1 - x}]}^{- 1}$
$= {(1 - x^{10})}^{- 1} (1 - x)$

$(1 + x^{10} + x^{20} + \dots .) (1 - x)$

Coeficient of $x^{401} = (- x) (x^{400})$

$= - 1 =$ Ans
Option (b)

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Similar questions

x^{401}

{(1 + x + x^{2} + \cdot \cdot \cdot + x^{9})}^{- 1}

(| x | < 1)

| x | < 1

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

x^{9}

(1 + x) (1 + x^{2}) (1 + x^{3}) . . . (1 + x^{100})

| x | < 1

x^{n}

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

x^{k} (0 \leq k \leq n)

E = 1 + (1 + x) + (1 + x)^{2} + . . . . + (1 + x)^{n}

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