If $n \in N$ and $\left(1+x+x^2+\ldots+x^p\right)^n=\sum_\limits{r=0}^{n p} a_r x^r$ if $0 \leq r \leq n p$ and $r$ is not a multiple of $p+1$, then ${ }^n C_0 a_r-{ }^n C_1 a_{r-1}+{ }^n C_2 a_{r-2}-\ldots+(-1)^r a_0{ }^n C_r$ is

JEE MATHEMATICS MCQ: BINOMIAL THEOREM

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Analyzing +4 | -1

n \in N

and

\left(1+x+x^2+\ldots+x^p\right)^n=\sum\limits_{{r=0}}^{n p} a_r x^r

0 \leq r \leq n p

and

r

is not a multiple of

p+1

, then

{ }^n C_0 a_r-{ }^n C_1 a_{r-1}+{ }^n C_2 a_{r-2}-\ldots+(-1)^r a_0{ }^n C_r

View answer and explanation

Correct option: C

Option C (0) is correct because when

r

is not a multiple of

p+1

, the coefficient of

x^r

in the expansion of

(1-x^{p+1})^n

is 0.

Practice All Chapter MCQs →

\left({ }^{10} C_0\right)^2-\left({ }^{10} C_1\right)^2+\ldots-\left({ }^{10} C_9\right)^2+\left({ }^{10} C_{10}\right)^2

a

d

\left( {n + 1} \right)

a{C_0} - \left( {a + d} \right){C_1} + \left( {a + 2d} \right){C_2} - ... + ...

(1-x)^{201}\left(1+x+x^2\right)^{200}

\left(\frac{1}{x}+2 x\right)^n

{x^5}

\frac{{1 + {x^2}}}{{1 + x}},\left| x \right| < 1,

\left(\begin{array}{c}2 n+1 \\ 0\end{array}\right)+\left(\begin{array}{c}2 n+1 \\ 3\end{array}\right)+\left(\begin{array}{c}2 n+1 \\ 6\end{array}\right)+\ldots=170

n