AP EAMCET · Maths · Sequences and Series
\(\sum_{k=1}^n k(k+1)(k+2) \ldots(k+r-1)=\)
- A \(\frac{n(n+1)(n+2) \ldots(n+r)}{r+1}\)
- B \(\frac{n(n+1)(n+2) \ldots(n+r-1)}{r}\)
- C \(\frac{\mathrm{n}(\mathrm{n}+1)(\mathrm{n}+2) \ldots(\mathrm{n}+\mathrm{r}+1)}{\mathrm{r}+1}\)
- D \(\frac{n(n+1)(n+2) \ldots 2 n}{2 n+1}\)
Answer & Solution
Correct Answer
(A) \(\frac{n(n+1)(n+2) \ldots(n+r)}{r+1}\)
Step-by-step Solution
Detailed explanation
\(\sum_{k=1}^n k(k+1)(k+2) \ldots(k+r-1) = \frac{n(n+1)(n+2) \ldots(n+r)}{r+1}\)
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