Question

\( \sum_{k=1}^{n} k(k+1)(k+2)...(k+r-1) = \)

• A. \( \frac{n(n+1)(n+2)...(n+r)}{r+1} \)
• B. \( \frac{n(n+1)(n+2)...(n+r-1)}{r} \)
• C. \( \frac{n(n+1)(n+2)...(n+r+1)}{r+1} \)
• D. \( \frac{n(n+1)(n+2)...2n}{2n+1} \)

Hint:

This is a standard summation. The general term is a product of \(r\) consecutive integers. Use the method of differences. Let \(T_k = k(k+1)...(k+r-1)\). Show \(T_k = \frac{1}{r+1} [k(k+1)...(k+r) - (k-1)k...(k+r-1)]\). Summing this from \(k=1\) to \(n\) gives a telescoping sum.

Answer 1

The Correct Option is A

Let \(T_k = k(k+1)(k+2).
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(k+r-1)\).
We aim to use the method of differences.
Consider the product of \(r+1\) terms: Let \(V_k = k(k+1)(k+2).
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(k+r-1)(k+r)\).
Then \(V_{k-1} = (k-1)k(k+1).
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(k+r-1)\).
Consider \(V_k - V_{k-1}\): \(V_k - V_{k-1} = k(k+1).
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(k+r-1)(k+r) - (k-1)k(k+1).
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(k+r-1)\) Factor out the common terms \(k(k+1).
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(k+r-1)\): \(V_k - V_{k-1} = k(k+1).
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(k+r-1) \left[ (k+r) - (k-1) \right]\) \(V_k - V_{k-1} = k(k+1).
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(k+r-1) (k+r-k+1)\) \(V_k - V_{k-1} = k(k+1).
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(k+r-1) (r+1)\) So, \(T_k = k(k+1).
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(k+r-1) = \frac{1}{r+1} (V_k - V_{k-1})\).
Now, we sum \(T_k\) from \(k=1\) to \(n\): \(S_n = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} \frac{1}{r+1} (V_k - V_{k-1})\) \(S_n = \frac{1}{r+1} \sum_{k=1}^{n} (V_k - V_{k-1})\).
The sum is a telescoping series: \( \sum_{k=1}^{n} (V_k - V_{k-1}) = (V_1 - V_0) + (V_2 - V_1) + \dots + (V_n - V_{n-1}) = V_n - V_0 \).
We have \(V_k = k(k+1).
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(k+r)\).
So, \(V_n = n(n+1).
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(n+r)\).
And \(V_0 = 0 \cdot (1) \cdot (2) \cdot \dots \cdot (r) = 0\).
Therefore, \(S_n = \frac{1}{r+1} (V_n - 0) = \frac{V_n}{r+1} = \frac{n(n+1)(n+2).
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(n+r)}{r+1}\).
\[ \boxed{\frac{n(n+1)(n+2).
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(n+r)}{r+1}} \]