Sum of Sequence of Binomial Coefficients by Sum of Powers of Integers
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Theorem
Let $n, k \in \Z_{\ge 0}$ be positive integers.
Let $S_k = \ds \sum_{i \mathop = 1}^n i^k$.
Then:
- $\ds \sum_{i \mathop = 1}^k \binom {k + 1} i S_i = \paren {n + 1}^{k + 1} - \paren {n + 1}$
Proof
Let $N$ be a positive integer.
Then:
\(\ds \paren {N + 1}^{k + 1} - N^{k + 1}\) | \(=\) | \(\ds \sum_{i \mathop = 0}^{k + 1} \binom {k + 1} i N^i - N^{k + 1}\) | Binomial Theorem: Integral Index | |||||||||||
\(\ds \) | \(=\) | \(\ds \binom {k + 1} 0 + \sum_{i \mathop = 1}^k \binom {k + 1} i N^i + \binom {k + 1} {k + 1} N^{k + 1} - N^{k + 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 1 + \sum_{i \mathop = 1}^k \binom {k + 1} i N^i\) | Definition of Binomial Coefficient |
Summing from $N = 1$ to $N = n$, we have on the left hand side:
\(\ds \sum_{N \mathop = 1}^n \paren {\paren {N + 1}^{k + 1} - N^{k + 1} }\) | \(=\) | \(\ds \paren {n + 1}^{k + 1} - 1^{k + 1}\) | Telescoping Series: Example 1 | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {n + 1}^{k + 1} - 1\) |
So, we have:
\(\ds \paren {n + 1}^{k + 1} - 1\) | \(=\) | \(\ds \sum_{N \mathop = 1}^n \paren {1 + \sum_{i \mathop = 1}^k \binom {k + 1} i N^i}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{N \mathop = 1}^n 1 + \sum_{i \mathop = 1}^k \binom {k + 1} i \sum_{N \mathop = 1}^n N^i\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds n + \sum_{i \mathop = 1}^k \binom {k + 1} i S_i\) |
giving:
- $\ds \sum_{i \mathop = 1}^k \binom {k + 1} i S_i = \paren {n + 1}^{k + 1} - \paren {n + 1}$
$\blacksquare$
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 19$: Sums of Powers of Positive Integers: $19.13$