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General Solution of a Differential Equation

If 1+x+x2+x...

Question

If $(1 + x + x^{2} + x^{3} + \dots \dots . x^{p})^{n} = a_{0} + a_{1^{X}} + a_{2} x^{2} + \dots . . + a_{n p} x^{n p}$ , then
$a_{0} + a_{1} + a_{2} + \dots . + a_{n p} =$

A

P^{n}

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B

(2 p + 1)^{n}

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C

(p + 1)^{n - 1}

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D

(p + 1)^{n}

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Solution

The correct option is D $(p + 1)^{n}$
Given,
$(1 + x^{2} + x^{3} + . . . . . . . . x^{p})^{n} = (a_{0} + a_{1} x + a_{2} x^{2} + . . . . . . . . . + a_{n p} x^{n} p)$

Substituting $x = 1$ in the above expression we get the sum of the binomial coefficients as
$\Rightarrow$ $(1 + x + x^{2} . . . x^{p})^{n}$
$= (1 + (1 + 1 + . . .1))^{n}$ (substituting x=1)
$= (1 + p (1))^{n}$
$= (1 + p)^{n}$
Hence answer is D

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Similar questions

(1 + x + x^{2} + . . . . . + x^{p})^{n} = a_{0} + a_{1} x + a_{2} x^{2} + . . . . . a_{n p} x^{n p}

(1) a_{0} + a_{1} + a_{2} + . . . . . + a_{n p} = (p + 1)^{n}

.

(2) a_{1} + 2 a_{2} + 3 a_{3} + . . . . + n p . a_{n p} = \frac{1}{2} n p (p + 1)^{n}

(1 - x)^{- n} = a_{0} + a_{1} x + a_{2} x^{2} + . . . . . . . . + a_{r} x^{r} + . . . . .

a_{0} + a_{1} + a_{2} + . . . . . . + a_{r}

\frac{1}{\sqrt{2 x + 1}} {{(1 + \sqrt{2 x + 1})}^{n} - {(1 - \sqrt{2 x + 1})}^{n}} = a_{0} + a_{1} x + a_{2} x^{2} + . . . . + a_{n} x^{n}

n

must be equal to

{(1 + x + x^{2})}^{n} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{2 n} x^{2 n},

find the values of
(i)

a_{0} + a_{1} + a_{2} + \dots . + a_{2 n}

a_{0} - a_{1} + a_{2} - a_{3} \dots . . + a_{2 n}

(1 + x + x^{2})^{20} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{40} x^{40},

then the value of

a_{0} + a_{1} + a_{2} + \dots + a_{19}

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