\sum_{k=1}^{n} k(k+1)(k+2) . . .(k+r-1)= [AP EAPCET 22 May 2025 S2]

Correct Answer is : n(n+1)(n+2)...(n+r)r+1

\frac{n(n+1)(n+2) . . .(n+r)}{r+1}

\frac{n(n+1)(n+2) . . .(n+r-1)}{r}

\frac{n(n+1)(n+2) . . .(n+r+1)}{r+1}

Sequences and Series

Section: Mathematics

Question : 4 of 25

Marks: +1, -0

∑

k=1

k(k+1)(k+2)...(k+r−1)=

[AP EAPCET 22 May 2025 S2]

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