JEE MATHEMATICS MCQ: BINOMIAL THEOREM

1 Medium

1 of 1

Analyzing +4 | -1

The coefficient of

x^r\{0 \leq r \leq(n-1)\}

in the expansion of

(x+2)^{n-1}+(x+2)^{n-2}(x+1)+(x+2)^{n-3}

(x+1)^2+\ldots+(x+1)^{n-1}

is equal to

View answer and explanation

Correct option: C

\left(2^{n-r}-1\right)^n C_r

Option C (

\left(2^{{n-r}}-1\right)^n C_r

) is correct because the given series can be simplified to

(x+2)^n - (x+1)^n

, and the coefficient of

x^r

in this expression is

\left(2^{{n-r}}-1\right)^n C_r

More Questions from BINOMIAL THEOREM

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