JEE MATHEMATICS MCQ: BINOMIAL THEOREM

1 Hard

1 of 1

Application +4 | -1

{S}_k=\sum\limits_{{r=1}}^n r^k

, then

\sum\limits_{{r=1}}^{k+1}\left(\begin{array}{c}k+1 \\ r\end{array}\right) {S}_{k+1 - r}=

View answer and explanation

Correct option: A

(n+1)^{k+1}-1

Option A (

(n+1)^{k+1}-1

) is correct because the given summation represents the expansion of

(n+1)^{k+1}

minus the term for

r=0

which is 1.

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