# Derive the Formula sum_(k=1)^nk^4=1/30(6n^5+15n^4+10n^3-n)?

## Not even sure where to start. Any guidance would be greatly appreciated. Thanks!

May 17, 2017

Consider first the following sum:

${\sum}_{k = 1}^{n} {\left(k + 1\right)}^{5} - {k}^{5}$

This is a sum of differences, and if we expand terms we can quickly establish the formula for this sum, as almost all the terms cancel:

${\sum}_{k = 1}^{n} {\left(k + 1\right)}^{5} - {k}^{5} = \left(\textcolor{red}{{2}^{5}} - {1}^{5}\right) + \left(\textcolor{b l u e}{{3}^{5}} - \textcolor{red}{{2}^{5}}\right) + \left(\textcolor{g r e e n}{{4}^{5}} - \textcolor{b l u e}{{3}^{5}}\right) + \left({5}^{5} - \textcolor{g r e e n}{{4}^{5}}\right) + \ldots + {\left(n + 1\right)}^{5} - {n}^{5}$

After cancelling we are left with:

$\setminus \setminus \setminus \setminus \setminus {\sum}_{k = 1}^{n} {\left(k + 1\right)}^{5} - {k}^{5} = - {1}^{5} + {\left(n + 1\right)}^{5}$
$\therefore {\sum}_{k = 1}^{n} {\left(k + 1\right)}^{5} - {k}^{5} = {\left(n + 1\right)}^{5} - 1 \setminus \setminus \ldots \left(\star\right)$

Now let us expand the terms on either side using the Binomial Theorem:

On the LHS we have:

${\left(k + 1\right)}^{5} - {k}^{5} = \left(1 + 5 k + 10 {k}^{2} + 10 {k}^{3} + 5 {k}^{4} + {k}^{5}\right) - {k}^{5}$
$\text{ } = 1 + 5 k + 10 {k}^{2} + 10 {k}^{3} + 5 {k}^{4}$

Similarly, on the RHS we have:

${\left(n + 1\right)}^{5} - 1 = \left(1 + 5 n + 10 {n}^{2} + 10 {n}^{3} + 5 {n}^{4} + {n}^{5}\right) - 1$
$\text{ } = 5 n + 10 {n}^{2} + 10 {n}^{3} + 5 {n}^{4} + {n}^{5}$

Combining these results into $\left(\star\right)$ we can write:

${\sum}_{k = 1}^{n} \left(1 + 5 k + 10 {k}^{2} + 10 {k}^{3} + 5 {k}^{4}\right) = 5 n + 10 {n}^{2} + 10 {n}^{3} + 5 {n}^{4} + {n}^{5}$

And so:

${\sum}_{k = 1}^{n} 1 + 5 {\sum}_{k = 1}^{n} k + 10 {\sum}_{k = 1}^{n} {k}^{2} + 10 {\sum}_{k = 1}^{n} {k}^{3} + 5 {\sum}_{k = 1}^{n} {k}^{4} = 5 n + 10 {n}^{2} + 10 {n}^{3} + 5 {n}^{4} + {n}^{5}$

We can now use the known standard summation formula:

${\sum}_{k = 1}^{n} \setminus k = \frac{1}{2} n \left(n + 1\right)$
${\sum}_{k = 1}^{n} {k}^{2} = \frac{1}{6} n \left(n + 1\right) \left(2 n + 1\right)$
${\sum}_{k = 1}^{n} {k}^{3} = \frac{1}{4} {n}^{2} {\left(n + 1\right)}^{2}$

And then we get:

$n + 5 \frac{1}{2} n \left(n + 1\right) + 10 \frac{1}{6} n \left(n + 1\right) \left(2 n + 1\right) + 10 \frac{1}{4} {n}^{2} {\left(n + 1\right)}^{2} + 5 {\sum}_{k = 1}^{n} {k}^{4} = 5 n + 10 {n}^{2} + 10 {n}^{3} + 5 {n}^{4} + {n}^{5}$

Multiplying by $6$ we get:

$6 n + 15 n \left(n + 1\right) + 10 n \left(n + 1\right) \left(2 n + 1\right) + 15 {n}^{2} {\left(n + 1\right)}^{2} + 30 {\sum}_{k = 1}^{n} {k}^{4} = 30 n + 60 {n}^{2} + 60 {n}^{3} + 30 {n}^{4} + 6 {n}^{5}$

Now multiply out the various brackets:

$6 n + 15 \left({n}^{2} + n\right) + 10 n \left(2 {n}^{2} + 3 n + 1\right) + 15 {n}^{2} \left({n}^{2} + 2 n + 1\right) + 30 {\sum}_{k = 1}^{n} {k}^{4} = 30 n + 60 {n}^{2} + 60 {n}^{3} + 30 {n}^{4} + 6 {n}^{5}$

$\therefore 6 n + 15 {n}^{2} + 15 n + 20 {n}^{3} + 30 {n}^{2} + 10 n + 15 {n}^{4} + 30 {n}^{3} + 15 {n}^{2} + 30 {\sum}_{k = 1}^{n} {k}^{4} = 30 n + 60 {n}^{2} + 60 {n}^{3} + 30 {n}^{4} + 6 {n}^{5}$

Collect terms on LHS and RHS:

$\therefore 31 n + 60 {n}^{2} + 50 {n}^{3} + 15 {n}^{4} + 30 {\sum}_{k = 1}^{n} {k}^{4} = 30 n + 60 {n}^{2} + 60 {n}^{3} + 30 {n}^{4} + 6 {n}^{5}$

Now isolate the sum by collection further terms:

$30 {\sum}_{k = 1}^{n} {k}^{4} = 30 n + 60 {n}^{2} + 60 {n}^{3} + 30 {n}^{4} + 6 {n}^{5} - 31 n - 60 {n}^{2} - 50 {n}^{3} - 15 {n}^{4}$

$\therefore 30 {\sum}_{k = 1}^{n} {k}^{4} = - n + 10 {n}^{3} + 15 {n}^{4} + 6 {n}^{5}$

$\therefore {\sum}_{k = 1}^{n} {k}^{4} = \frac{1}{30} \left(6 {n}^{5} + 15 {n}^{4} + 10 {n}^{3} - n\right)$ QED

May 17, 2017

Examine sequences of differences to find the formula...

#### Explanation:

Since the sum is of polynomial terms of degree $4$, the formula for the sum will be a polynomial of degree $5$.

If the terms of a sequence are given by a polynomial of degree $n$, then forming the sequence of differences, then of differences of differences $n$ times will result in a constant sequence.

So the sequence of sums, being a polynomial of degree $5$ will reduce to a constant sequence when we take differences $5$ times.

To prepare, let us write down the first few powers of $4$ (six terms will be just enough):

$1 , 16 , 81 , 256 , 625 , 1296$

Now write the sequence of the first $6$ sums:

$\textcolor{b l u e}{1} , 17 , 98 , 354 , 979 , 2275$

Next write the sequence of differences between pairs of successive terms:

$\textcolor{b l u e}{16} , 81 , 256 , 625 , 1296$

Write down the sequence of differences of those differences:

$\textcolor{b l u e}{65} , 175 , 369 , 671$

Write down the sequence of differences of those differences:

$\textcolor{b l u e}{110} , 194 , 302$

Write down the sequence of differences of those differences:

$\textcolor{b l u e}{84} , 108$

Write down the sequence of differences of those differences:

$\textcolor{b l u e}{24}$

We can now take the initial terms of each of these sequences as coefficients for a formula for the $n$th term of the sequence of sums:

s_n = color(blue)(1)/(0!)+color(blue)(16)/(1!)(n-1)+color(blue)(65)/(2!)(n-1)(n-2)+color(blue)(110)/(3!)(n-1)(n-2)(n-3)+color(blue)(84)/(4!)(n-1)(n-2)(n-3)(n-4)+color(blue)(24)/(5!)(n-1)(n-2)(n-3)(n-4)(n-5)

$\textcolor{w h i t e}{{s}_{n}} = 1 + 16 n - 16 + \frac{65}{2} {n}^{2} - \frac{195}{2} n + 65 + \frac{55}{3} {n}^{3} - 110 {n}^{2} + \frac{605}{3} n - 110 + \frac{7}{2} {n}^{4} - 35 {n}^{3} + \frac{245}{2} {n}^{2} - 175 n + 84 + \frac{1}{5} {n}^{5} - 3 {n}^{4} + 17 {n}^{3} - 45 {n}^{2} + \frac{274}{5} n - 24$

$\textcolor{w h i t e}{{s}_{n}} = \frac{1}{5} {n}^{5} + \frac{1}{2} {n}^{4} + \frac{1}{3} {n}^{3} - \frac{1}{30} n$

$\textcolor{w h i t e}{{s}_{n}} = \frac{1}{30} \left(6 {n}^{5} + 15 {n}^{4} + 10 {n}^{3} - n\right)$

May 18, 2017

#### Explanation:

Making the hypothesis

${\sum}_{k = 1}^{n} {k}^{4} = {c}_{5} {n}^{5} + {c}_{4} {n}^{4} + {c}_{3} {n}^{3} + {c}_{2} {n}^{2} + {c}_{1} n + {c}_{0}$

we have

$\left\{\begin{matrix}k = 1 \to {c}_{0} + {c}_{1} + {c}_{2} + {c}_{3} + {c}_{4} + {c}_{5} = 1 \\ k = 2 \to {c}_{0} + 2 {c}_{1} + 4 {c}_{2} + 8 {c}_{3} + 16 {c}_{4} + 32 {c}_{5} = 17 \\ k = 3 \to {c}_{0} + 3 {c}_{1} + 9 {c}_{2} + 27 {c}_{3} + 81 {c}_{4} + 243 {c}_{5} = 98 \\ k = 4 \to {c}_{0} + 4 {c}_{1} + 16 {c}_{2} + 64 {c}_{3} + 256 {c}_{4} + 1024 {c}_{5} = 354 \\ k = 5 \to {c}_{0} + 5 {c}_{1} + 25 {c}_{2} + 125 {c}_{3} + 625 {c}_{4} + 3125 {c}_{5} = 979 \\ k = 6 \to {c}_{0} + 6 {c}_{1} + 36 {c}_{2} + 216 {c}_{3} + 1296 {c}_{4} + 7776 {c}_{5} = 2275\end{matrix}\right.$

now solving for ${c}_{0} , {c}_{1} , {c}_{2} , {c}_{3} , {c}_{4} , {c}_{5}$ we obtain

${c}_{0} = 0 , {c}_{1} = - \frac{1}{30} , {c}_{2} = 0 , {c}_{3} = \frac{1}{3} , {c}_{4} = \frac{1}{2} , {c}_{5} = \frac{1}{5}$

or

${\sum}_{k = 1}^{n} {k}^{4} = - \frac{n}{30} + {n}^{3} / 3 + {n}^{4} / 2 + {n}^{5} / 5 = \frac{1}{30} n \left(1 + n\right) \left(1 + 2 n\right) \left(3 n \left(1 + n\right) - 1\right)$