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November 6th, 2017, 10:56 AM  #1 
Newbie Joined: Nov 2017 From: Latvia Posts: 2 Thanks: 0  I need to understand and help for a proof
Find a closed formula for the sum f(n) = 1^4 + 2^4 + 3^4 + . . . + n^4

November 6th, 2017, 11:04 AM  #2 
Senior Member Joined: Aug 2012 Posts: 2,043 Thanks: 584  
November 7th, 2017, 12:06 AM  #3 
Global Moderator Joined: Dec 2006 Posts: 19,708 Thanks: 1805 
The formula is n(n + 1)(2n + 1)(3n² + 3n  1)/30, which can be proved by mathematical induction.

December 1st, 2017, 03:42 AM  #4 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 
You can derive that formula using "Newton's divided difference formula": Setting $\displaystyle f(n)= 1+ 2^4+ 3^4+ \cdot\cdot\cdot+ n^4$ with f(0)= 0, f(1)= 1, f(2)= 17, f(3)= 98, f(4)= 354, f(5)= 989, f(6)= 2275. The "first differences" are $\displaystyle \Delta f(0)= f(1) f(0)= 1$ $\displaystyle \Delta f(1)= f(2) f(1)= 16$ $\displaystyle \Delta f(2)= f(3) f(2)= 81$ $\displaystyle \Delta f(3)= f(4) f(3)= 256$ $\displaystyle \Delta f(4)= f(5) f(4)= 625$ $\displaystyle \Delta f(5)= f(6) f(5)= 1296$ (Those are, of course, the fourth powers.) The "second differences" are $\displaystyle \Delta^2 f(0)= 16 1= 15$ $\displaystyle \Delta^2 f(1)= 81 16= 65$ $\displaystyle \Delta^2 f(2)= 256 81= 175$ $\displaystyle \Delta^2 f(3)= 625 256= 369$ $\displaystyle \Delta^2 f(4)= 1296 625= 671$ The "third differences" are $\displaystyle \Delta^3 f(0)= 65 15= 50$ $\displaystyle \Delta^3 f(1)= 175 65= 110$ $\displaystyle \Delta^3 f(2)= 369 175= 194$ $\displaystyle \Delta^3 f(3)= 671 369= 302$ The "fourth differences" are $\displaystyle \Delta^4(0)= 110 50= 60$ $\displaystyle \Delta^4(1)= 194 110= 84$ $\displaystyle \Delta^4(2)= 302 194= 108$ The "fifth differences" are $\displaystyle \Delta^5(0)= 84 60= 24$ $\displaystyle \Delta^5(1)= 108 84= 24$ The "fifth differences" are all the same (which you can verify by extending the xvalues larger than 6 and calculating further differences) so all succeeding differences are 0. "Newton's divided difference method" says that $\displaystyle f(x)= f(0)+ \Delta f(0)x \frac{\Delta^2 f(0)}{2}x(x 1)+ \frac{\Delta^3 f(0)}{3!}x(x 1)(x 2)+ \cdot\cdot\cdot$. Note the similarities to (and differences from) Taylor's series. Here, that gives $\displaystyle f(x)= 0+ x+ \frac{15}{2}x(x 1)+ \frac{50}{6}x(x 1)(x 2)+ \frac{60}{24}x(x 1)(x 2)(x 3)+ \frac{24}{120}x(x 1)(x 2)(x 3)(x 4)$ 
December 1st, 2017, 06:24 AM  #5 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 
Yet another way: knowing, perhaps from "divided differences" as above, that the sum, to n, of kth powers is a k+ 1 degree polynomial, we can write this as fifth degree polynomial in n: an^5+ bn^4+ cn^3+ dn^2+ en+ f. Taking n= 0, f= 0. Taking n= 1, a+ b+ c+ d+ e+ f= a+ b+ c+ d+ e= 1. Taking n= 2. 32a+ 16b+ 8c+ 4d+ 2e= 17. Taking n= 3, 242a+ 81b+ 27c+ 9d+ 3e= 98. Taking n= 4, 1024a+ 256b+ 64c+ 16d+ 4e= 354. Taking n= 5, 3125a+ 625b+ 125c+ 25d+ 5e= 989. That is 6 linear equations to solve for a, b, c, d, e, and f. 
December 1st, 2017, 07:52 AM  #6 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,445 Thanks: 2499 Math Focus: Mainly analysis and algebra 
\begin{align*} n(n1) &= 1 \\ \sum_{k=1}^n \big( k  (k1) \big) &= \sum_{k=1}^n 1 \\ \big(n  \cancel{(n1)}\big) + \big(\cancel{(n1)}  \cancel{(n2)}\big) + \ldots + \big(\cancel{2}  \cancel{1}\big) + \big(\cancel{1}  0\big) &= \sum_{k=1}^n 1 \\ \sum_{k=1}^n 1 &= n \\[8pt] \hline n^2(n1)^2 &= n^2  (n^2  2n + 1) = 2n1 \\ \sum_{k=1}^n \big( k^2  (k1)^2 \big) &= \sum_{k=1}^n (2k1) \\ \big(n^2  \cancel{(n1)^2}\big) + \big(\cancel{(n1)^2}  \cancel{(n2)^2}\big) + \ldots + \big(\cancel{2^2}  \cancel{1^2}\big) + \big(\cancel{1^2}  0\big) &= \sum_{k=1}^n 2k  \sum_{k=1}^n 1 \\ n^2 &= 2\sum_{k=1}^n k  \sum_{k=1}^n 1 \\ \sum_{k=1}^n k &= \frac12 \left(n^2 + \sum_{k=1}^n 1\right) \\ &= \frac12\left(n^2 + n\right) = \frac12n(n+1) \\[8pt] \hline n^3(n1)^3 &= n^3  (n^3  3n^2 + 3n  1) = 3n^2  3n + 1 \\ \sum_{k=1}^n \big( k^3  (k1)^3 \big) &= \sum_{k=1}^n (3k^2  3k + 1) \\ \big(n^3  \cancel{(n1)^3}\big) + \big(\cancel{(n1)^3}  \cancel{(n2)^3}\big) + \ldots + \big(\cancel{2^3}  \cancel{1^3}\big) + \big(\cancel{1^3}  0^3\big) &= \sum_{k=1}^n 3n^2  \sum_{k=1}^n 3n + \sum_{k=1}^n 1 \\ n^3 &= 3\sum_{k=1}^n k^2  3\sum_{k=1}^n k + \sum_{k=1}^n 1 \\ \sum_{k=1}^n k^2 &= \frac13 \left(n^3 + 3\sum_{k=1}^n k  \sum_{k=1}^n 1\right) \\ &= \frac12\left(n^3 + \frac32n(n+1)  n\right) = \frac16n(2n+1)(n+1) \\[8pt] \hline & \vdots \end{align*} etc. 
December 9th, 2017, 07:08 PM  #7 
Senior Member Joined: Jul 2012 From: DFW Area Posts: 626 Thanks: 90 Math Focus: Electrical Engineering Applications 
I think that I have another way of determining the coefficients using the differentials as given in post #4. It involves two (possible) properties of Pascal's Triangle (PT). I will start with $n=1$, with the inverted triangle below (row 0 is just the terms): $\displaystyle \large {\begin{array}{c c} & n & 1 & & 2 & & 3 & & 4 & & 5 & & 6 & & 7 \\ row \\ 0 & & 1 & & 17 & & 98 & & 354 & & 979 & & 2275 & & 4676 \\ 1 & & & 16 & & 81 & & 256 & & 625 & & 1296 & & 2401 \\ 2 & & & & 65 & & 175 & & 369 & & 671 & & 1105 \\ 3 & & & & & 110 & & 195 & & 302 & & 434 \\ 4 & & & & & & 84 & & 108 & & 132 \\ 5 & & & & & & & 24 & & 24 \end{array}}$ As stated in earlier posts, the number of subtractions to get to a constant is 5, so it is a fifth order polynomial that we are looking for, $a \cdot n^5+b \cdot n^4+c \cdot n^3+d \cdot n^2+e \cdot n+f$.  To get $a$: The constant is 24 so divide by 5!=120 $\displaystyle a=\large \frac{24}{5!}= \frac{1}{5}$ The factorial is the number of subtractions needed to get to the constant. Now that we have $a$, we could subtract off $a \cdot n^5$ from each term and do another triangle until we get a constant and divide the constant by 4!. But instead let's use the triangle that we already have.  To get $b$: First calculate the following sum using the value for $a$: $\displaystyle \large s = \frac{1}{5} \left (1 \cdot 5^5  4 \cdot 4^5 + 6 \cdot 3^5  4 \cdot 2^5 + 1 \cdot 1^5 \right)=72$ Then: $\displaystyle \large b=\frac{8472}{4!}=\frac{12}{24}=\frac{1}{2}$ with the value of 84 coming from the first term in row 4 of the inverted triangle above. The power of the numbers 5 down to 1 is 5 (since it is $a \cdot n^5$), and the coefficients come from the fourth row of PT, starting with a + and then alternating signs.  To get $c$: First calculate the following sum using the values for $a$ and $b$: $\displaystyle \large s = \frac{1}{5} \left (1 \cdot 4^5  3 \cdot 3^5 + 3 \cdot 2^5  1 \cdot 1^5 \right) + \frac{1}{2} \left (1 \cdot 4^4  3 \cdot 3^4 + 3 \cdot 2^4  1 \cdot 1^4 \right) = 108$ Then: $\displaystyle \large c=\frac{110108}{3!}=\frac{2}{6}=\frac{1}{3}$ Notice that the coefficients are for row 3 of PT.  To get $d$: First calculate the following sum using the values for $a$, $b$, and $c$: $\displaystyle \large s = \frac{1}{5} \left (1 \cdot 3^5  2 \cdot 2^5 + 1 \cdot 1^5 \right) + \frac{1}{2} \left (1 \cdot 3^4  2 \cdot 2^4 + 1 \cdot 1^4 \right) + \frac{1}{3} \left (1 \cdot 3^3  2 \cdot 2^3 + 1 \cdot 1^3 \right)= 65$ Then: $\displaystyle \large d=\frac{6565}{2!}=0$  To get $e$: First calculate the following sum using the values for $a$, $b$, $c$, and $d$: $\displaystyle \large s = \frac{1}{5} \left (1 \cdot 2^5  1 \cdot 1^5 \right) + \frac{1}{2} \left (1 \cdot 2^4  1 \cdot 1^4 \right) + \frac{1}{3} \left (1 \cdot 2^3  1 \cdot 1^3 \right )+ 0 \cdot \left (1 \cdot 2^2  1 \cdot 1^2 \right)= 16 \frac{1}{30}$ Then: $\displaystyle \large e=\frac{1616 \frac{1}{30}}{1!}=\frac{1}{30}$  To get f: $\displaystyle \large 1=\frac{1}{5}+\frac{1}{2}+\frac{1}{3}\frac{1}{30}+f$ $\displaystyle \large f=0$  So the polynomial is: $\displaystyle \large \frac{1}{5} \cdot n^5 + \frac{1}{2} \cdot n^4 + \frac{1}{3} \cdot n^3  \frac{1}{30} \cdot n$ Has anyone seen this method? I will post a new thread try to explain it. 

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