1^4+2^4+3^4+..........+n^4 =n(n+1)(2n+1)+(3n^2+3n-1)/30. (whole divided by 30)

to prove the given result by mathematical induction method, first we will show that it is true for n=1
then we will assume that it is true for n = k, and with the help of assumption we will try to prove that the given result is true for n = k+1.
step I:
P(1) = 1
RHS = 1*2*(2+1)*(3+3-1)30=2*3*530=1
thus the given equation holds for n =1
step II:
let us assume that given equation holds for n = k
therefore
14+24+.......+k4=k(k+1)(2k+1)(3k2+3k-1)30   ............(1)
step III:
now let us find the sum for n = (k+1)
14+24+......+k4+(k+1)4=k(k+1)(2k+1)(3k2+3k-1)30+(k+1)4=(k+1)30.[k(2k+1)(3k2+3k-1)+30(k+1)3]=(k+1)30.6k4+6k3-2k2+3k3+3k2-k+30.{k3+3k2+3k+1}=(k+1)30.6k4+9k3+k2-k+30k3+90k2+90k+30=(k+1)30.6k4+39k3+91k2+89k+30
now we will find the factor of g(k)=6k4+39k3+91k2+89k+30
g(-2)=6*(-2)4+39*(-2)3+91*(-2)2+89*(-2)+30=96-312+364-178+30=490-490=0
thus (k+2) is a factor of g(k).
6k4+39k3+91k2+89k+30= 6k3 (k+2)+27k2(k+2)+37k(k+2)+15(k+2)=6k4+12k3+27k3+54k2+37k2+74k+15k+30g(k)=(k+2)(6k3+27k2+37k+15)
now k = -3/2 is a factor of g(k).
thus by the factorization we can write:
g(k)=(k+2)(2k+3)(3k2+9k+5)
therefore
14+24+34+...........+k4+(k+1)4=(k+1)30.(k+2)(2k+3)(3k2+9k+5)=(k+1)30.{(k+1)+1}.{2.(k+1)+1}.{3.(k+1)2+3(k+1)-1}
thus the given equation also holds for n = k+1.
thus it is true for all integer values of n.

hope this helps you

 

  • -10
What are you looking for?