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[eqn]\sum_{n=1}^x n = \frac{1}{2}(x)(x+1)[/eqn]
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>>8349372
[eqn]\sum_{n=1}^x n^{2} = \frac{1}{6}(x)(x+1)(2x+1)[/eqn]
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>>8349373
[eqn]\sum_{n=1}^x n^{3} = \frac{1}{4}(x)(x)(x+1)(x+1)[/eqn]
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>>8349375
[eqn]\sum_{n=1}^x n^{4} = \frac{1}{30}(x)(x+1)(2x+1)(3x^{2}+3x-1)[/eqn]
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>>8349377
[eqn]\sum_{n=1}^x n^{5} = \frac{1}{12}(x)(x)(x+1)(x+1)(2x^{2}+2x-1)[/eqn]
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>>8349378
[eqn]\sum_{n=1}^x n^{6} = \frac{1}{42}(x)(x+1)(2x+1)(3x^{4}+6x^{3}-3x+1)[/eqn]
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>>8349380
[eqn]\sum_{n=1}^x n^{7} = \frac{1}{24}(x)(x)(x+1)(x+1)(3x^{4}+6x^{3}-x^{2}-4x+2)[/eqn]
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>>8349383
[eqn]\sum_{n=1}^x n^{8} = \frac{1}{90}(x)(x+1)(2x+1)(5x^{6}+15x^{5}+5x^{4}-15x^{3}-x^{2}+9x-3)[/eqn]
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>>8349385
[eqn]\sum_{n=1}^x n^{9} = \frac{1}{20}(x)(x)(x+1)(x+1)(2x^{6}+6x^{5}+x^{4}-8x^{3}+x^{2}+6x-3)[/eqn]
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>>8349387
[eqn] \sum_{n=1}^x n^{10} = \frac{1}{66}(x)(x+1)(2x+1)(3x^{8}+12x^{7}+8x^{6}-18x^{5}-10x^{4}+24x^{3}+2x^{2}-15x+5) [/eqn]
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>>8349388
[eqn] \sum_{n=1}^x n^{11} = \frac{1}{24}(x)(x)(x+1)(x+1)(2x^{8}+8x^{7}+4x^{6}-16x^{5}-5x^{4}+26x^{3}-3x^{2}-20x+10) [/eqn]
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>>8349390
[eqn] \sum_{n=1}^x n^{12} = \frac{1}{2730}(x)(x+1)(2x+1)(105x^{10}+525x^{9}+525x^{8}-1050x^{7}-1190x^{6}+2310x^{5}+1420x^{4}-3285x^{3}-287x^{2}+2073x-691) [/eqn]
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>>8349391
[eqn] \sum_{n=1}^x n^{13} = \frac{1}{420}(x)(x)(x+1)(x+1)(30x^{10}+150x^{9}+125x^{8}-400x^{7}-326x^{6}+1052x^{5}+367x^{4}-1786x^{3}+202x^{2}+1382x-691) [/eqn]
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>>8349393
[eqn] \sum_{n=1}^x n^{14} = \frac{1}{90}(x)(x+1)(2x+1)(3x^{12}+18x^{11}+24x^{10}-45x^{9}-81x^{8}+144x^{7}+182x^{6}-345x^{5}-217x^{4}+498x^{3}+44x^{2}-315x+105) [/eqn]
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>>8349394
[eqn] \sum_{n=1}^x n^{15} = \frac{1}{48}(x)(x)(x+1)(x+1)(3x^{12}+18x^{11}+21x^{10}-60x^{9}-83x^{8}+226x^{7}+203x^{6}-632x^{5}-226x^{4}+1084x^{3}-122x^{2}-840x+420) [/eqn]
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>>8349395
[eqn] \sum_{n=1}^x n^{16} = \frac{1}{510}(x)(x+1)(2x+1)(15x^{14}+105x^{13}+175x^{12}-315x^{11}-805x^{10}+1365x^{9}+2775x^{8}-4845x^{7}-6275x^{6}+11835x^{5}+7485x^{4}-17145x^{3}-1519x^{2}+10851x-3617) [/eqn]
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>>8349398
[eqn] \sum_{n=1}^x n^{17} = \frac{1}{180}(x)(x)(x+1)(x+1)(10x^{14}+70x^{13}+105x^{12}-280x^{11}-565x^{10}+1410x^{9}+2165x^{8}-5740x^{7}-5271x^{6}+16282x^{5}+5857x^{4}-27996x^{3}+3147x^{2}+21702x-10851) [/eqn]
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>>8349372
Is there any pattern relating the coefficients to the power of n? Seems pretty interesting
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>>8349402
Faulhaber's formula
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>>8349372
proof
S = 1+2+3+...+(n-1)+n
S = n+(n-1)+...+3+2+1
2S =(1+n)+(2+n-1)+...+(n-1+2)+(n+1) = n*(n+1)
S= 1/2 * n * (n+1)
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[math] \sum_{k=1}^n k^p = \sum_{j=0}^p {{ p+1 } \choose j} \dfrac{ (-1)^{j} \, B_j } {p+1} n^{p+1-j},\qquad \mbox{where}~B_1 = -\frac{1}{2} [/math]
where [math] B_j [/math] are the triggering Bernulle numbers (-1)(1/2), (-2)(-1/12), etc.
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>>8349418
[math] \sum_{j=0}^\infty B_j \dfrac {x^j} {j!} = \dfrac { x }{ e^x - 1 } e^{n \, x} [/math]
where you may want to use
[math] \dfrac { 1 }{ 1 - e^x } = -1 + \dfrac {1} {2}\,x - \dfrac {1} {12}\,x^2 + \dots [/math]
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>>8349440
Some observations:
[math] \sum_{k=1}^n k^p = \dfrac {n^{p+1}}{p+1} - \sum_{j=0}^{p-1} { p \choose j } \zeta(-j) n^{p-j} [/math]
where
[math] \zeta(j) = \frac{ 1 } { (j-1)! } \int_0^{\infty} \dfrac { 1 } { e ^ x - 1 } \, x^{j-1} \, dx [/math]
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>>8349447
of what use is your formula since e does not exist
Thread posts: 25
Thread images: 19
Thread images: 19