# Sum $sum_{(k_1, k_2, k_3): k_1+k_2+k_3=K, ,, n_1+n_2+n_3=N}k_1^{n_1}times k_2^{n_2} times k_3^{n_3}$

Mathematics Asked on January 5, 2022

Let $$k_1, k_2, k_3$$ be natural non-negative numbers such that $$k_1+k_2+k_3=K$$. Let $$n_1, n_2, n_3 in {0, ldots, N}$$ and such that $$n_1+n_2+n_3=N$$.

Calculate

$$S=sum_{(k_1, k_2, k_3): k_1+k_2+k_3=K, ,, n_1+n_2+n_3=N}k_1^{n_1}times k_2^{n_2} times k_3^{n_3}$$

My attempt:
I am thinking on representing this sum as a chain of sums over each summand $$k_j$$. For example, the interior sum would be:
$$sum_{k_3=0}^{K-k_1-k_2}k_3^{n_3}.$$
Using Sums of p-th powers formula we can get $$sum_{k_3=0}^{K-k_1-k_2}k_3^{n_3}=frac{B_{n_3+1}(K-k_1-k_2+1)-B_{n_3+1}}{n_3}.$$ So, the sum $$S$$ would be represented as a product of these ratios with Bernoulli numbers $$B_n$$.

Is there a better way on computing/estimating from above sum $$S$$?

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