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Question Number 6902 by Tawakalitu. last updated on 02/Aug/16

1^6 + 3^6 + 5^6 + ... + (2n − 1)^6 = B Solve for B.

$$\mathrm{1}^{\mathrm{6}} \:+\:\mathrm{3}^{\mathrm{6}} \:+\:\mathrm{5}^{\mathrm{6}} \:+\:...\:+\:\left(\mathrm{2}{n}\:−\:\mathrm{1}\right)^{\mathrm{6}} \:\:=\:\:{B} \\ $$$$ \\ $$$${Solve}\:{for}\:{B}. \\ $$

Commented by Yozzii last updated on 03/Aug/16

Method of undetermined coefficients Let s(n)=Σ_(r=−3) ^n (2r−1)^6 (n∈Z) where s(n)=an^7 +bn^6 +cn^5 +dn^4 +en^3 +fn^2 +gn+h (a,b,c,d,e,f,g,h are 8 real constants). I chose a polynomial of degree 7=6+1 since the following expressions show that the polynomial should have a power one greater than that of the general term. Σ_(i=1) ^n i^1 =((n(n+1))/2)=((n^2 +n)/2) (2=1+1) Σ_(i=1) ^n i^2 =((n(n+1)(2n+1))/6)=((2n^3 +3n^2 +n)/6) (3=2+1) Σ_(i=1) ^n i^3 =((n^2 (n+1)^2 )/4)=((n^4 +2n^3 +n^2 )/4) (4=3+1) −−−−−−−−−−−−−−−−−−−−−−−−−− ∴ n=−3: s(−3)=(−7)^6 =a(−3)^7 +b(−3)^6 +c(−3)^5 +d(−3)^4 +e(−3)^3 +f(−3)^2 +g(−3)+h (1)117649=−2187a+729b−243c+81d−27e+9f−3g+h ∴ n=−2: s(−2)=(−7)^6 +(−5)^6 =−128a+64b−32c+16d−8e+4f−2g+h (2) 133274=−128a+64b−32c+16d−8e+4f−2g+h ∴ n=−1: s(−1)=(−7)^6 +(−5)^6 +(−3)^6 =−a+b−c+d−e+f−g+h (3) 134003=−a+b−c+d−e+f−g+h ∴ n=0: s(0)=0+0+0+0+0+0+0+h (4) 134004=h ∴ n=1: s(1)=a+b+c+d+e+f+g+h (5) 134005=a+b+c+d+e+f+g+h ∴n=2: s(2)=128a+64b+32c+16d+8e+4f+2g+h (6) 134734=128a+64b+32c+16d+8e+4f+2g+h ∴ n=3: s(3)=2187a+729b+243c+81d+27e+9f+3g+h (7) 150359=2187a+729b+243c+81d+27e+9f+3g+h ∴n=4: s(4)=16384a+4096b+1024c+256d+64e+16f+4g+h (8) 268008=16384a+4096b+1024c+256d+64e+16f+4g+h −−−−−−−−−−−−−−−−−−−−−−−−−−− We have from (4), h=134004=s(0) −−−−−−−−−−−−−−−−−−−−−− After solving for the real unknown coefficients, you get a=((64)/7),c=−16,e=((28)/3),g=((−31)/(21)), h=134004, b=d=f=0. −−−−−−−−−−−−−−−−−−−−−−− ∴ S(n)=Σ_(i=1) ^n (2i−1)^6 =s(n)−s(0)=(n/(21))(192n^6 −336n^4 +196n^2 −31) S(n)=(n/(21))(2n−1)(2n+1)(48n^4 −72n^2 +31) Or, B=((n(2n−1)(2n+1)(48^4 −72n^2 +31))/(21))

$$\boldsymbol{{M}}{ethod}\:{of}\:{undetermined}\:{coefficients} \\ $$$${Let}\:{s}\left({n}\right)=\underset{{r}=−\mathrm{3}} {\overset{{n}} {\sum}}\left(\mathrm{2}{r}−\mathrm{1}\right)^{\mathrm{6}} \:\:\left({n}\in\mathbb{Z}\right) \\ $$$${where}\:{s}\left({n}\right)={an}^{\mathrm{7}} +{bn}^{\mathrm{6}} +{cn}^{\mathrm{5}} +{dn}^{\mathrm{4}} +{en}^{\mathrm{3}} +{fn}^{\mathrm{2}} +{gn}+{h} \\ $$$$\left({a},{b},{c},{d},{e},{f},{g},{h}\:{are}\:\mathrm{8}\:{real}\:{constants}\right). \\ $$$${I}\:{chose}\:{a}\:{polynomial}\:{of}\:{degree}\:\mathrm{7}=\mathrm{6}+\mathrm{1} \\ $$$${since}\:{the}\:{following}\:{expressions}\:{show}\:{that} \\ $$$${the}\:{polynomial}\:{should}\:{have}\:{a}\:{power}\:{one} \\ $$$${greater}\:{than}\:{that}\:{of}\:{the}\:{general}\:{term}. \\ $$$$\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}{i}^{\mathrm{1}} =\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}=\frac{{n}^{\mathrm{2}} +{n}}{\mathrm{2}}\:\:\left(\mathrm{2}=\mathrm{1}+\mathrm{1}\right) \\ $$$$\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}{i}^{\mathrm{2}} =\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}}=\frac{\mathrm{2}{n}^{\mathrm{3}} +\mathrm{3}{n}^{\mathrm{2}} +{n}}{\mathrm{6}}\:\left(\mathrm{3}=\mathrm{2}+\mathrm{1}\right) \\ $$$$\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}{i}^{\mathrm{3}} =\frac{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}}=\frac{{n}^{\mathrm{4}} +\mathrm{2}{n}^{\mathrm{3}} +{n}^{\mathrm{2}} }{\mathrm{4}}\:\:\left(\mathrm{4}=\mathrm{3}+\mathrm{1}\right) \\ $$$$−−−−−−−−−−−−−−−−−−−−−−−−−− \\ $$$$\therefore\:{n}=−\mathrm{3}:\:{s}\left(−\mathrm{3}\right)=\left(−\mathrm{7}\right)^{\mathrm{6}} ={a}\left(−\mathrm{3}\right)^{\mathrm{7}} +{b}\left(−\mathrm{3}\right)^{\mathrm{6}} +{c}\left(−\mathrm{3}\right)^{\mathrm{5}} +{d}\left(−\mathrm{3}\right)^{\mathrm{4}} +{e}\left(−\mathrm{3}\right)^{\mathrm{3}} +{f}\left(−\mathrm{3}\right)^{\mathrm{2}} +{g}\left(−\mathrm{3}\right)+{h} \\ $$$$\left(\mathrm{1}\right)\mathrm{117649}=−\mathrm{2187}{a}+\mathrm{729}{b}−\mathrm{243}{c}+\mathrm{81}{d}−\mathrm{27}{e}+\mathrm{9}{f}−\mathrm{3}{g}+{h} \\ $$$$ \\ $$$$\therefore\:{n}=−\mathrm{2}:\:{s}\left(−\mathrm{2}\right)=\left(−\mathrm{7}\right)^{\mathrm{6}} +\left(−\mathrm{5}\right)^{\mathrm{6}} =−\mathrm{128}{a}+\mathrm{64}{b}−\mathrm{32}{c}+\mathrm{16}{d}−\mathrm{8}{e}+\mathrm{4}{f}−\mathrm{2}{g}+{h} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{133274}=−\mathrm{128}{a}+\mathrm{64}{b}−\mathrm{32}{c}+\mathrm{16}{d}−\mathrm{8}{e}+\mathrm{4}{f}−\mathrm{2}{g}+{h} \\ $$$$ \\ $$$$\therefore\:{n}=−\mathrm{1}:\:{s}\left(−\mathrm{1}\right)=\left(−\mathrm{7}\right)^{\mathrm{6}} +\left(−\mathrm{5}\right)^{\mathrm{6}} +\left(−\mathrm{3}\right)^{\mathrm{6}} =−{a}+{b}−{c}+{d}−{e}+{f}−{g}+{h} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{134003}=−{a}+{b}−{c}+{d}−{e}+{f}−{g}+{h} \\ $$$$ \\ $$$$\therefore\:{n}=\mathrm{0}:\:{s}\left(\mathrm{0}\right)=\mathrm{0}+\mathrm{0}+\mathrm{0}+\mathrm{0}+\mathrm{0}+\mathrm{0}+\mathrm{0}+{h} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{134004}={h} \\ $$$$ \\ $$$$\therefore\:{n}=\mathrm{1}:\:{s}\left(\mathrm{1}\right)={a}+{b}+{c}+{d}+{e}+{f}+{g}+{h} \\ $$$$\left(\mathrm{5}\right)\:\mathrm{134005}={a}+{b}+{c}+{d}+{e}+{f}+{g}+{h} \\ $$$$ \\ $$$$\therefore{n}=\mathrm{2}:\:{s}\left(\mathrm{2}\right)=\mathrm{128}{a}+\mathrm{64}{b}+\mathrm{32}{c}+\mathrm{16}{d}+\mathrm{8}{e}+\mathrm{4}{f}+\mathrm{2}{g}+{h} \\ $$$$\left(\mathrm{6}\right)\:\mathrm{134734}=\mathrm{128}{a}+\mathrm{64}{b}+\mathrm{32}{c}+\mathrm{16}{d}+\mathrm{8}{e}+\mathrm{4}{f}+\mathrm{2}{g}+{h} \\ $$$$ \\ $$$$\therefore\:{n}=\mathrm{3}:\:{s}\left(\mathrm{3}\right)=\mathrm{2187}{a}+\mathrm{729}{b}+\mathrm{243}{c}+\mathrm{81}{d}+\mathrm{27}{e}+\mathrm{9}{f}+\mathrm{3}{g}+{h} \\ $$$$\left(\mathrm{7}\right)\:\mathrm{150359}=\mathrm{2187}{a}+\mathrm{729}{b}+\mathrm{243}{c}+\mathrm{81}{d}+\mathrm{27}{e}+\mathrm{9}{f}+\mathrm{3}{g}+{h} \\ $$$$ \\ $$$$\therefore{n}=\mathrm{4}:\:{s}\left(\mathrm{4}\right)=\mathrm{16384}{a}+\mathrm{4096}{b}+\mathrm{1024}{c}+\mathrm{256}{d}+\mathrm{64}{e}+\mathrm{16}{f}+\mathrm{4}{g}+{h} \\ $$$$\left(\mathrm{8}\right)\:\mathrm{268008}=\mathrm{16384}{a}+\mathrm{4096}{b}+\mathrm{1024}{c}+\mathrm{256}{d}+\mathrm{64}{e}+\mathrm{16}{f}+\mathrm{4}{g}+{h} \\ $$$$−−−−−−−−−−−−−−−−−−−−−−−−−−− \\ $$$${We}\:{have}\:{from}\:\left(\mathrm{4}\right),\:{h}=\mathrm{134004}={s}\left(\mathrm{0}\right) \\ $$$$−−−−−−−−−−−−−−−−−−−−−− \\ $$$${After}\:{solving}\:{for}\:{the}\:{real}\:{unknown}\:{coefficients}, \\ $$$${you}\:{get}\:{a}=\frac{\mathrm{64}}{\mathrm{7}},{c}=−\mathrm{16},{e}=\frac{\mathrm{28}}{\mathrm{3}},{g}=\frac{−\mathrm{31}}{\mathrm{21}},\:{h}=\mathrm{134004},\:{b}={d}={f}=\mathrm{0}. \\ $$$$−−−−−−−−−−−−−−−−−−−−−−− \\ $$$$\therefore\:{S}\left({n}\right)=\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\mathrm{2}{i}−\mathrm{1}\right)^{\mathrm{6}} ={s}\left({n}\right)−{s}\left(\mathrm{0}\right)=\frac{{n}}{\mathrm{21}}\left(\mathrm{192}{n}^{\mathrm{6}} −\mathrm{336}{n}^{\mathrm{4}} +\mathrm{196}{n}^{\mathrm{2}} −\mathrm{31}\right) \\ $$$${S}\left({n}\right)=\frac{{n}}{\mathrm{21}}\left(\mathrm{2}{n}−\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{48}{n}^{\mathrm{4}} −\mathrm{72}{n}^{\mathrm{2}} +\mathrm{31}\right) \\ $$$${Or},\:{B}=\frac{{n}\left(\mathrm{2}{n}−\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{48}^{\mathrm{4}} −\mathrm{72}{n}^{\mathrm{2}} +\mathrm{31}\right)}{\mathrm{21}} \\ $$

Commented by Tawakalitu. last updated on 03/Aug/16

Interesting... wow ... thanks so much for your help

$${Interesting}...\:{wow}\:...\:{thanks}\:{so}\:{much}\:{for}\:{your}\:{help} \\ $$

Answered by Yozzii last updated on 03/Aug/16

Answer in comments

$${Answer}\:{in}\:{comments} \\ $$

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