Question

How many summands are in $\sum _{i=i_0}^l\sum _{j=j_0}^m\sum _{k=k_0}^n{k}^i{j}^k$?

Short answer

Ans: The total number of summands in$\sum _{i=i_0}^l\sum _{j=j_0}^m\sum _{k=k_0}^n{k}^i{j}^k$is$\left(k_0-n+1\right)\times \left(j_0-m+1\right)$

Step-by-step solution

Step 1. Given information: 

$\sum _{i=i_0}^l\sum _{j=j_0}^m\sum _{k=k_0}^n{k}^i{j}^k$

Step 2. First Summation:

$$\begin{gathered} \sum _{i=i_0}^{i=l}\sum _{j=j_0}^{j=m}\sum _{k=k_0}^{k=n}{k}^i{j}^k=\sum _{i=i_0}^{i=l}\left(\sum _{j=j_0}^{j=m}\sum _{k=k_0}^{k=n}{k}^i{j}^k\right) \\ =\sum _{i=i_0}^{i=l}\left(\sum _{j=j_0}^{j=m}\left(\sum _{k=k_0}^{k=n}{k}^i{j}^k\right)\right) \end{gathered}$$

The summation $\sum _{k=k_0}^{k=n}{k}^i{j}^k$can be expanded by replacing k with $k_0$and rewriting each term when k ends at n.

Hence, the number of terms in the summation $\sum _{k=k_0}^{k=n}{k}^i{j}^k$is $k_0-n+1$

Step 3. Second Summation:

Similarly, when expanded $\sum _{j=j_0}^{j=m}\left(\sum _{k=k_0}^{k=n}{k}^i{j}^k\right)$by replacing j from $j_0$to m, the number of $\sum _{k=k_0}^{k=n}{k}^i{j}^k$terms obtained is $j_0-m+1$

Hence, the total number of summands of $\sum _{j=j_0}^{j=m}\left(\sum _{k=k_0}^{k=n}{k}^i{j}^k\right)$is $\left(k_0-n+1\right)\times \left(j_0-m+1\right)$

Step 4. Third Summation: 

Similarly, when expanded $\sum _{i=i_0}^{i=l}\left(\sum _{j=j_0}^{j=m}\left(\sum _{k=k_0}^{k=n}{k}^i{j}^k\right)\right)$by replacing i from $i_0$to l, the number of $\sum _{j=j_0}^{j=m}\left(\sum _{k=k_0}^{k=n}{k}^i{j}^k\right)$terms obtained is $i_0-l+1$

The total number of summands in $\sum _{i=b_0}^{i=l}\sum _{j=j_0}^{j=m}\sum _{k=k_0}^{k=n}{k}^i{j}^k$is $\left(i_0-l+1\right)\times \left(k_0-n+1\right)\times \left(j_0-m+1\right)$