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

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

Thanks!

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

Thanks!

##### 3 Answers

Consider first the following sum:

# sum_(k=1)^n (k+1)^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 (k+1)^5-k^5 = (color(red)(2^5)-1^5) + (color(blue)(3^5)-color(red)(2^5))+ (color(green)(4^5)-color(blue)(3^5)) + (5^5-color(green)(4^5))+... + (n+1)^5-n^5 #

After cancelling we are left with:

# \ \ \ \ \ sum_(k=1)^n (k+1)^5-k^5 = -1^5 + (n+1)^5 #

# :. sum_(k=1)^n (k+1)^5-k^5 = (n+1)^5 -1 \ \ ... (star)#

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

On the LHS we have:

# (k+1)^5-k^5 = (1+5k+10k^2+10k^3 +5k^4+k^5) -k^5 #

# " " = 1+5k+10k^2+10k^3 +5k^4 #

Similarly, on the RHS we have:

# (n+1)^5 -1 = (1+5n+10n^2+10n^3 +5n^4+n^5) -1 #

# " " = 5n+10n^2+10n^3 +5n^4+n^5 #

Combining these results into

# sum_(k=1)^n (1+5k+10k^2+10k^3 +5k^4) = 5n+10n^2+10n^3 +5n^4+n^5 #

And so:

# sum_(k=1)^n 1 + 5sum_(k=1)^nk + 10sum_(k=1)^nk^2 + 10sum_(k=1)^nk^3 + 5sum_(k=1)^nk^4 = 5n + 10n^2 + 10n^3 + 5n^4 + n^5 #

We can now use the known standard summation formula:

# sum_(k=1)^n \ k = 1/2n(n+1) #

# sum_(k=1)^n k^2 = 1/6n(n+1)(2n+1) #

# sum_(k=1)^n k^3 = 1/4n^2(n+1)^2 #

And then we get:

# n + 5 1/2 n(n+1) + 10 1/6 n(n+1)(2n+1) + 10 1/4 n^2(n+1)^2 + 5sum_(k=1)^nk^4 = 5n + 10n^2 + 10n^3 + 5n^4 + n^5 #

Multiplying by

# 6n + 15n(n+1) + 10n(n+1)(2n+1) + 15n^2(n+1)^2 + 30sum_(k=1)^nk^4 = 30n + 60n^2 + 60n^3 + 30n^4 + 6n^5 #

Now multiply out the various brackets:

# 6n + 15(n^2+n) + 10n(2n^2+3n+1) + 15n^2(n^2+2n+1) + 30sum_(k=1)^nk^4 = 30n + 60n^2 + 60n^3 + 30n^4 + 6n^5 #

# :. 6n + 15n^2+15n + 20n^3+30n^2+10n + 15n^4+30n^3+15n^2 + 30sum_(k=1)^nk^4 = 30n + 60n^2 + 60n^3 + 30n^4 + 6n^5 #

Collect terms on LHS and RHS:

# :. 31n + 60n^2 + 50n^3 + 15n^4 + 30sum_(k=1)^nk^4 = 30n + 60n^2 + 60n^3 + 30n^4 + 6n^5 #

Now isolate the sum by collection further terms:

# 30sum_(k=1)^nk^4 = 30n + 60n^2 + 60n^3 + 30n^4 + 6n^5 - 31n - 60n^2 - 50n^3 - 15n^4 #

# :. 30sum_(k=1)^nk^4 = -n + 10n^3 + 15n^4 + 6n^5 #

# :. sum_(k=1)^nk^4 = 1/30(6n^5+ 15n^4+10n^3 -n) # QED

#### Answer:

Examine sequences of differences to find the formula...

#### Explanation:

Since the sum is of polynomial terms of degree

If the terms of a sequence are given by a polynomial of degree

So the sequence of sums, being a polynomial of degree

To prepare, let us write down the first few powers of

#1, 16, 81, 256, 625, 1296#

Now write the sequence of the first

#color(blue)(1), 17, 98, 354, 979, 2275#

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

#color(blue)(16), 81, 256, 625, 1296#

Write down the sequence of differences of those differences:

#color(blue)(65), 175, 369, 671#

Write down the sequence of differences of those differences:

#color(blue)(110), 194, 302#

Write down the sequence of differences of those differences:

#color(blue)(84), 108#

Write down the sequence of differences of those differences:

#color(blue)(24)#

We can now take the initial terms of each of these sequences as coefficients for a formula for the

#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)#

#color(white)(s_n) = 1+16n-16+65/2n^2-195/2n+65+55/3n^3-110n^2+605/3n-110+7/2n^4-35n^3+245/2n^2-175n+84+1/5n^5-3n^4+17n^3-45n^2+274/5n-24#

#color(white)(s_n) = 1/5n^5 + 1/2n^4 + 1/3n^3 - 1/30n#

#color(white)(s_n) = 1/30(6n^5 +15n^4+10n^3-n)#

#### Answer:

The number-crunching answer.

#### Explanation:

Making the hypothesis

we have

now solving for

or