JEE MATHEMATICS MCQ: BINOMIAL THEOREM

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Application +4 | -1

For any positive integers

m, n

with

n \geq m

, let

\left(\begin{array}{l}n \\ m\end{array}\right)={ }^n C_m

. Then,

\left(\begin{array}{l}n \\ m\end{array}\right)+\left(\begin{array}{c}n-1 \\ m\end{array}\right)+\left(\begin{array}{c}n-2 \\ m\end{array}\right)+\ldots+\left(\begin{array}{l}m \\ m\end{array}\right)

is equal to

View answer and explanation

Correct option: C

{ }^{n+1} C_{m+1}

Option C (

{ }^{n+1} C_{m+1}

) is correct because the sum of binomial coefficients can be simplified using the identity

{ }^n C_r + { }^n C_{r-1} = { }^{n+1} C_r

repeatedly.

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