Answer to Question #173434 in Calculus for Justin Suson

Question #173434

Derive a formula for

n

i= i3 a telescoping sum with terms f(I) = i4

1
Expert's answer
2021-03-31T01:58:45-0400

TO DERIVE :

let,

S=i=1ni3S=\displaystyle\sum_{i=1}^ni^3 ............(1)


We start off the proof with the "telescoping sum" (or the collapsing sum).


i=1n[(i+1)4i4]\displaystyle\sum_{i=1}^n [(i+1)^4-i^4] SOLVING IN GENERAL FORM


=(2414)+(3424)+(4434)+...........+................+(n+1)4n4+...........+................+(n)4(n1)4=(n+1)414=(n=1)41=((n+1)2)21=(n2+2n+1)21=(\cancel{2^4}-1^4)+(\cancel{3^4}-\cancel{2^4})+(4^4-\cancel{3^4})\\\\+...........+................+(n+1)^4-\cancel{n^4}\\+...........+................+ \cancel{(n)^4}-\cancel{(n-1)^4}\\ =(n+1)^4-1^4\\ =(n=1)^4-1\\ =((n+1)^2)^2-1\\ =(n^2+2n+1)^2-1


i=1n[(i+1)4i4]\displaystyle\sum_{i=1}^n [(i+1)^4-i^4] =n4+4n3+6n2+4n+1=n^4+4n^3+6n^2+4n+1 .......................(2)


now,


i=1n[(i+1)4i4]\displaystyle\sum_{i=1}^n [(i+1)^4-i^4] SOLVING IN terms of i



=i=1n[(i4+4i3+6i2+4i+1)i4]=i=1n[(4i3+6i2+4i+1)]=4i=1ni3+6i=1ni2+4i=1ni+i=1n1​ =\displaystyle\sum_{i=1}^n[(i^4+4i^3+6i^2+4i+1)-i^4]\\ =\displaystyle\sum_{i=1}^n [(4i^3+6i^2+4i+1)] \\ =4\displaystyle\sum_{i=1}^ni^3+6\displaystyle\sum_{i=1}^ni^2+4\displaystyle\sum_{i=1}^ni+\displaystyle\sum_{i=1}^n1\\

from (1) ..further solving above...


=4S+6(n(n+1)(2n+1)6)+4(n(n+1)2)+n=4S+6({n(n+1)(2n+1)\over6})+4({n(n+1)\over2})+n

=4S+2n3+n2+2n2+n+2n2+3n=4S+2n3+5n2+4n=4S+2n^3+n^2+2n^2+n+2n^2+3n\\ =4S+2n^3+5n^2+4n ................(3)


THUS; equating (2) and (3);


n4+4n3+6n2+4n=4S+2n3+5n2+4nn4+2n3+n2=4Sn2(n2+2n+1)=4Sn^4+4n^3+6n^2+4n=4S+2n^3+5n^2+4n\\ n^4+2n^3+n^2=4S\\ n^2(n^2+2n+1)=4S


4S=n2(n+1)2S=n2(n+1)2224S=n^2(n+1)^2\\ S={n^2(n+1)^2\over2^2}


S=[n(n+1)2]2\boxed{S=[{n(n+1)\over2}]^2}

hence drived.






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