I can't find the sum of n^4 using the telescoping method.

[The Student]

I think that my instructor wants us to find the sum of n^4 by only using the telescoping method and knowing the sums of n^3, n^2 and n. Is this possible, or should I use a different method to find this sum?

 

[soroban]

Hello, The Student!

My instructor wants us to find the sum of n^4 by only using the telescoping method
and knowing the sums of n^3, n^2 and n.
Is this possible, or should I use a different method to find this sum?

I believe I know this method . . . It's long and messy.

Note that: .\(\displaystyle k^5 - (k-1)^5 \:=\:5k^4 - 10k^3 + 10k^2 - 5k + 1\)

Let \(\displaystyle k \,=\,n,n\!-\!1,n\!-\!2,\cdots, 3,2,1\) and "stack" the equations.

\(\displaystyle \begin{array}{ccccccccccccccc}&n^5 &-& (n-1)^5 &=& 5n^4 &-& 10n^3 &+& 10n^2 &-& 5n &+& 1 \\
&(n-1)^5 &-& (n-2)^5 &=& 5(n-1)^4 &-& 10(n-1)^3 &+& 10(n-1)^2 &-& 5(n-1) &+& 1 \\
&(n-2)^5 &-& (n-3)^5 &=& 5(n-2)^4 &-& 10(n-2)^3 &+& 10(n-2)^2 &-& 5(n-2) &+& 1 \\
&\vdots&&\vdots && \vdots && \vdots && \vdots && \vdots && \vdots \\
&3^5 &-& 2^5 &=& 5\cdot3^4 &-& 10\cdot3^3 &+& 10\cdot3^2 &-& 5\cdot3 &+& 1 \\
&2^5 &-& 1^5 &=& 5\cdot2^4 &-& 10\cdot2^3 &+& 10\cdot2^2 &-& 5\cdot2 &+& 1 \\
&1^5 &-& 0^5 &=& 5\cdot1^4 &-& 10\cdot1^3 &+& 10\cdot1^2 &-& 5\cdot1 &+& 1 \\ \hline
\text{Add:} && n^5 && = & 5\sum k^4 &-& 10\sum k^3 &+& 10\sum k^2 &-& 5\sum k &+& \sum 1 \end{array}\)

We have: .\(\displaystyle n^5 \;=\;5\sum k^4 - 10\cdot\dfrac{n^2(n+1)^2}{4} + 10\cdot\dfrac{n(n+1)(2n+1)}{6} -5\cdot\dfrac{n(n+1)}{2} + n \)

. . . . . . . .\(\displaystyle n^5 \;=\;5\sum k^4 - \dfrac{5n^2(n+1)^2}{2} + \dfrac{5n(n+1)(2n+1)}{3} - \dfrac{5n(n+1)}{2}+n \)

. . . . \(\displaystyle 5\sum k^4 \;=\;n^5 - n + \dfrac{5n^2(n+1)^2}{2} - \dfrac{5n(n+1)(2n+1)}{3} + \dfrac{5n(n+1)}{2}\)


Multiply by 6: .\(\displaystyle \displaystyle 30\sum^n_{k=1} k^4 \;=\;6(n^4-1) + 15n^2(n+1)^2 - 10n(n+1)(2n+1) + 15n(n+1)\)

. . . . . . . . . . .\(\displaystyle \displaystyle30\sum^n_{k=1} k^4 \;=\;6(n-1)(n+1)(n^2+1) + 15n^2(n+1)^2 - 10n(n+1)(2n+1) + 15n(n+1)\)

Factor: . . . . .\(\displaystyle \displaystyle30\sum^n_{k=1} k^4 \;=\;n(n+1)\bigg[6(n-1)(n^2+1) + 15n(n+1) - 10(2n+1) + 15\bigg]\)

. . . . . . . . . . .\(\displaystyle \displaystyle30\sum^n_{k=1} k^4 \;=\;n(n+1)(6n^3 + 9n^2 + n - 1)\)


Therefore: . . . . \(\displaystyle \displaystyle\sum^n_{k=1}k^4 \;=\;\dfrac{n(n+1)(6n^3 + 9n^2 + n - 1)}{30}\)

 

[The Student]

Thank-you, I would have never thought of that way. I actually just found out another way. The sum of ((ak+1)^(m+1) - ak^(m+1)) where ak+1 = k, ak = (k - 1) and m = 4, from k = 1 to n. We can have the sum of k, from k = 1 to n, equal (k - 1) by adjusting the start and end of the sum to go from k = 0 to n - 1, and when we subtract this summation series from the other summation of (k - 1)^5 which is from k = 1 to n, we are left with n^5. Now we can write the above all over again; only this time we can leave the summation in the telescoping formula ((ak+1)^(m+1) - ak^(m+1)) and equate it to n^5, and then we expand the polynomials inside the summation brackets. The two k^5's cancel out and the sum of the expanded variables including k^4 are left. We sum each variable, and we isolate the summation of k^4. So we put the summation of k^4 on one side and equate it to the rest that we know already as the degrees left are n^3, n^2 and n which is given in the question. So, I just thought that I would throw this out there in case you forgot this or didn't know it. I always take from this website, so I am glad to give back when I can.