Can you derive formula for sum of 4th powers of first n natural numbers

Algebra Level 4

Let

$S=1^4 + 2^4+3^4+4^4+...............+n^4.$

Then which of the following expressions gives the value of $S?$

\frac{n^5+10(\frac{n(n+1)}{2})^2-\frac{5n(n+1)(2n+1)}{3}+\frac{5n(n+1)}{2}-n}{5}

\frac{n^5+\frac{5n^2(n-1)^2}{4}+(3n+5)^3-n}{28}

It is not possible to express

S

in terms of n.

\frac{n^2(n+1)^2}{4}+

sum of product of

n

natural numbers taken 2 at a time.

1 solution

Aman Sharma
Oct 24, 2014

We know that:- $x^5-(x-1)^5=5x^4-10x^3+10x^2-5x+1$ Now put x=1,2,3,4,,.....n-1,n one by one in above equation:- $(1)^5-(0)^5=5(1)^4-10(1)^3+10(1)^2-5(1)+1$ $(2)^5-(1)^5=5(2)^4-10(2)^3+10(2)^2-5(2)+1$ $(3)^5-(2)^5=5(3)^4-10(3)^3+10(3)^2-5(3)+1$ $..................................$ $..................................$ $(n-1)^5-(n-2)^5=5(n-1)^4-10(n-1)^3+10(n-1)^2-5(n-1)+1$ $(n)^5-(n-1)^5=5(n)^4-10(n)^3+10(n)^2-5(n)+1$ Adding all these equations it is clear that all terms in LHS will cancel out except $n^5)$ so:- $n^5=5(1^4+2^4+3^4+...+n^5)-10(1^3+2^3+3^3+....+n^3)+10(1^2+2^2+3^2+....+n^2)-5(1+2+3+.....+n)+(1+1+1+..+n times)..............(1)$ Now we know that:- $1^3+2^3+3^3+....+n^3=(\frac{n(n+1)}{2})^2$ \ $1^3+2^2+3^2+....+n^2=\frac{n(n+1)(2n+1)}{6}$ $1+2+3+.....+n=\frac{n(n+1)}{2}$ Putting all these values in equation (1) and rearranging gives the required expression....

options are not clearly observable

nikhil jaiswal - 6 years, 7 months ago

Yes. You're right...I lost my chance due to this :(

Krishna Ar - 6 years, 7 months ago

@Krishna Ar someone has reported my problem and sorry i am not well in latex coding...i read somewhere that you are a moderator.....can you please edit.....multiple choise opetions