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Question: Let $k$ and $n$ be the positive integers and $S_k = 1^k + 2^k + 3^k + .... + n^k$. Then $^{m+1}C_1S_...

Let kk and nn be the positive integers and Sk=1k+2k+3k+....+nkS_k = 1^k + 2^k + 3^k + .... + n^k. Then m+1C1S1+m+1C2S2+m+1C3S3+.....m+1CmSm^{m+1}C_1S_1 + ^{m+1}C_2S_2 + ^{m+1}C_3S_3 + ..... ^{m+1}C_mS_m is equal to :

A

(n+1)m+1(n+1)^{m+1}

B

(n+1)m−n(n+1)^m - n

C

(n+1)m+1−1(n+1)^{m+1} - 1

D

(n+1)m+1−n−1(n+1)^{m+1} - n - 1

Answer

(n+1)m+1−n−1(n+1)^{m+1} - n - 1

Explanation

Solution

We start with

Sk=1k+2k+⋯+nk.S_k = 1^k + 2^k + \cdots + n^k.

The given sum is

∑k=1m(m+1k)Sk=∑k=1m(m+1k)∑r=1nrk.\sum_{k=1}^{m} \binom{m+1}{k} S_k = \sum_{k=1}^{m} \binom{m+1}{k} \sum_{r=1}^{n} r^k.

Interchange summation:

=∑r=1n∑k=1m(m+1k)rk.= \sum_{r=1}^{n} \sum_{k=1}^{m} \binom{m+1}{k} r^k.

Recognize that

∑k=0m+1(m+1k)rk=(1+r)m+1.\sum_{k=0}^{m+1} \binom{m+1}{k} r^k = (1+r)^{m+1}.

Subtract the terms corresponding to k=0k=0 and k=m+1k=m+1 (since our summation is from 1 to mm):

∑k=1m(m+1k)rk=(1+r)m+1−1−rm+1.\sum_{k=1}^{m} \binom{m+1}{k} r^k = (1+r)^{m+1} - 1 - r^{m+1}.

Thus, our original sum becomes:

∑r=1n[(1+r)m+1−1−rm+1]=∑r=1n(1+r)m+1−n−∑r=1nrm+1.\sum_{r=1}^{n}\left[(1+r)^{m+1} - 1 - r^{m+1}\right] = \sum_{r=1}^{n}(1+r)^{m+1} - n - \sum_{r=1}^{n} r^{m+1}.

Shift the index in the first summation by setting s=r+1s = r+1:

∑r=1n(1+r)m+1=∑s=2n+1sm+1=∑s=1n+1sm+1−1.\sum_{r=1}^{n} (1+r)^{m+1} = \sum_{s=2}^{n+1} s^{m+1} = \sum_{s=1}^{n+1} s^{m+1} - 1.

Substitute back:

=(∑s=1n+1sm+1−1)−n−∑r=1nrm+1.= \left(\sum_{s=1}^{n+1} s^{m+1} - 1\right) - n - \sum_{r=1}^{n} r^{m+1}.

Notice the telescoping nature:

∑s=1n+1sm+1−∑r=1nrm+1=(n+1)m+1.\sum_{s=1}^{n+1} s^{m+1} - \sum_{r=1}^{n} r^{m+1} = (n+1)^{m+1}.

Thus, the expression simplifies to:

(n+1)m+1−1−n=(n+1)m+1−n−1.(n+1)^{m+1} - 1 - n = (n+1)^{m+1} - n - 1.