Maverick848
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Why is the sum of the first n cubes the square of the sum of the first n integers? How would you go about proving this?
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- Thread starter Maverick848
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Here's a method you may like:
The proof can generalize formulas for sums of the form \(\displaystyle k^{m}\)
\(\displaystyle \sum_{k=1}^{n}{(k+1)^{4}-k^{4}}=(2^{4}-1^{4})+(3^{4}-2^{4})+(4^{4}-3^{4})+.........+[(n+1)^{4}-n^{4}]=(n+1)^{4}-1\) [1]
Because all of the other terms in this telescoping sum cancel out.
It is true that \(\displaystyle (k+1)^{4}-k^{4}=4k^{3}+6k{2}+4k+1\)
Plug this into the left side of [1]:
\(\displaystyle \sum_{k=1}^{n}{4k^{3}+6k{2}+4k+1}=(n+1)^{4}-1=n^{4}+4n^{3}+6n^{2}+4n\)
Now split the properties of summations:
\(\displaystyle \sum_{k=1}^{n}{4k^{3}+6k^{2}+4k+1}=4\sum_{k=1}^{n}{k^{3}}+6\sum_{k=1}^{n}{k^{2}}+4\sum_{k=1}^{n}{k}+\sum_{k=1}^{n}{1}\)
\(\displaystyle \sum_{k=1}^{n}{4k^{3}+6k^{2}+4k+1}=4\sum_{k=1}^{n}{k^{3}}+6(\frac{n(n+1)(2n+1)}{6})+4(\frac{n(n+1)}{2})+n\)
\(\displaystyle \sum_{k=1}^{n}{4k^{3}+6k^{2}+4k+1}=4\sum_{k=1}^{n}{k^{3}}+n(n+1)(2n+1))+2n(n+1)+n\)
\(\displaystyle 4\sum_{k=1}^{n}{k^{3}}=n^{4}+4n^{3}+6n^{2}+4n-n(n+1)(2n+1)-2n(n+1)-n\)
\(\displaystyle 4\sum_{k=1}^{n}{k^{3}}=n^{2}(n+1)^{2}\)
\(\displaystyle \sum_{k=1}^{n}{k^{3}}=\frac{n^{2}(n+1)^{2}}{4}=(\frac{(n(n+1)}{2})^{2}\)
Another interesting formula for the sum of the cubes is done with binomial
identities. Do a google search.
\(\displaystyle C(n+1,2)+6C(n+1,3)+6C(n+1,4)\).
Give it a try. Let n be, say, 25
\(\displaystyle C(26,2)+6C(26,3)+6C(26,4)=105,625\)
Use the formula:
\(\displaystyle \frac{(25)^{2}(25+1)^{2}}{4}=105,625\)
This is derived using binomial identities.
There are numerous methods for deriving the formulae for the sums of the powers. It's an interesting mathematical field.
The proof can generalize formulas for sums of the form \(\displaystyle k^{m}\)
\(\displaystyle \sum_{k=1}^{n}{(k+1)^{4}-k^{4}}=(2^{4}-1^{4})+(3^{4}-2^{4})+(4^{4}-3^{4})+.........+[(n+1)^{4}-n^{4}]=(n+1)^{4}-1\) [1]
Because all of the other terms in this telescoping sum cancel out.
It is true that \(\displaystyle (k+1)^{4}-k^{4}=4k^{3}+6k{2}+4k+1\)
Plug this into the left side of [1]:
\(\displaystyle \sum_{k=1}^{n}{4k^{3}+6k{2}+4k+1}=(n+1)^{4}-1=n^{4}+4n^{3}+6n^{2}+4n\)
Now split the properties of summations:
\(\displaystyle \sum_{k=1}^{n}{4k^{3}+6k^{2}+4k+1}=4\sum_{k=1}^{n}{k^{3}}+6\sum_{k=1}^{n}{k^{2}}+4\sum_{k=1}^{n}{k}+\sum_{k=1}^{n}{1}\)
\(\displaystyle \sum_{k=1}^{n}{4k^{3}+6k^{2}+4k+1}=4\sum_{k=1}^{n}{k^{3}}+6(\frac{n(n+1)(2n+1)}{6})+4(\frac{n(n+1)}{2})+n\)
\(\displaystyle \sum_{k=1}^{n}{4k^{3}+6k^{2}+4k+1}=4\sum_{k=1}^{n}{k^{3}}+n(n+1)(2n+1))+2n(n+1)+n\)
\(\displaystyle 4\sum_{k=1}^{n}{k^{3}}=n^{4}+4n^{3}+6n^{2}+4n-n(n+1)(2n+1)-2n(n+1)-n\)
\(\displaystyle 4\sum_{k=1}^{n}{k^{3}}=n^{2}(n+1)^{2}\)
\(\displaystyle \sum_{k=1}^{n}{k^{3}}=\frac{n^{2}(n+1)^{2}}{4}=(\frac{(n(n+1)}{2})^{2}\)
Another interesting formula for the sum of the cubes is done with binomial
identities. Do a google search.
\(\displaystyle C(n+1,2)+6C(n+1,3)+6C(n+1,4)\).
Give it a try. Let n be, say, 25
\(\displaystyle C(26,2)+6C(26,3)+6C(26,4)=105,625\)
Use the formula:
\(\displaystyle \frac{(25)^{2}(25+1)^{2}}{4}=105,625\)
This is derived using binomial identities.
There are numerous methods for deriving the formulae for the sums of the powers. It's an interesting mathematical field.
Here are a few pthers to ponder.
*-- The sum of the first n positive odd integers, 1+3+5+.....+n, is n^2.
*-- The sum of the first n positive even integers, 2+4+6+.....+n, is n(n + 1)
*-- The sum of the first n positive integers, 1+2+3+.....+n, is n(n + 1)/2 or (n^2 + n)/2
*--The sum of the squares of the first n integers, 1^2+2^2+3^2+.....+n^2, is n(n+1)(2n+1)/6
*--The sum of the cubes of the first n integers, 1^3+2^3+3^3+.....+n^3, is [(n^2 + n)/2]^2
*--n^3 is always the sum of n odd integers beginning with (n^2 - n + 1) and ending with (n^2 + n - 1).
*--The sum of (n + 1) consecutive squares, beginning with the square of n(2n + 1), is equal to the sum of the squares of the next n consecutive squares, e.g., for n = 1, 3^2 + 4^2 = 5^2; for n = 2, 10^2 + 11^2 + 12^2 = 13^2 + 14^2; etc.
*--The product of any n consecutive positive integers is divisible by the product of the first n postitive integers.
*--Every even number equal to or greater than 6 is the sum of two odd primes.
*--Every odd number equal to or greater than 9 is the sum of three odd primes.
*--Every sum of two consecutive odd primes is the product of three integers, all greater than one.
*--The sum of two cubes is never a cube.
*--The nth term of the series 1 + 2 + 4 + 8 + 16 + .........= 2^(n-1).
*--The sum of the first n terms of the series 1 + 2 + 4 + 8 + 16 + ......, or 1, 3, 7, 15, ----- is 2^n - 1.
*--The sum of the first n cubes is [n(n-1)/2]^3.
*--The sum of the first n terms of the series 1 + x + x^2 + x^3 + ------x^(n-1) = (x^n - 1)/(x - 1).
*--1^3 + 2^3 + 3^3 + 4^3 + -----n^3 = (1 + 2 + 3 + 4 + -----n)^2.
*--The product of any n consecutive integers is divisible by the product of the first n integers.
*-- The sum of the first n positive odd integers, 1+3+5+.....+n, is n^2.
*-- The sum of the first n positive even integers, 2+4+6+.....+n, is n(n + 1)
*-- The sum of the first n positive integers, 1+2+3+.....+n, is n(n + 1)/2 or (n^2 + n)/2
*--The sum of the squares of the first n integers, 1^2+2^2+3^2+.....+n^2, is n(n+1)(2n+1)/6
*--The sum of the cubes of the first n integers, 1^3+2^3+3^3+.....+n^3, is [(n^2 + n)/2]^2
*--n^3 is always the sum of n odd integers beginning with (n^2 - n + 1) and ending with (n^2 + n - 1).
*--The sum of (n + 1) consecutive squares, beginning with the square of n(2n + 1), is equal to the sum of the squares of the next n consecutive squares, e.g., for n = 1, 3^2 + 4^2 = 5^2; for n = 2, 10^2 + 11^2 + 12^2 = 13^2 + 14^2; etc.
*--The product of any n consecutive positive integers is divisible by the product of the first n postitive integers.
*--Every even number equal to or greater than 6 is the sum of two odd primes.
*--Every odd number equal to or greater than 9 is the sum of three odd primes.
*--Every sum of two consecutive odd primes is the product of three integers, all greater than one.
*--The sum of two cubes is never a cube.
*--The nth term of the series 1 + 2 + 4 + 8 + 16 + .........= 2^(n-1).
*--The sum of the first n terms of the series 1 + 2 + 4 + 8 + 16 + ......, or 1, 3, 7, 15, ----- is 2^n - 1.
*--The sum of the first n cubes is [n(n-1)/2]^3.
*--The sum of the first n terms of the series 1 + x + x^2 + x^3 + ------x^(n-1) = (x^n - 1)/(x - 1).
*--1^3 + 2^3 + 3^3 + 4^3 + -----n^3 = (1 + 2 + 3 + 4 + -----n)^2.
*--The product of any n consecutive integers is divisible by the product of the first n integers.