# 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 = $\frac{1*2*\left(2+1\right)*\left(3+3-1\right)}{30}=\frac{2*3*5}{30}=1$
thus the given equation holds for n =1
step II:
let us assume that given equation holds for n = k
therefore

step III:
now let us find the sum for n = (k+1)
${1}^{4}+{2}^{4}+......+{k}^{4}+\left(k+1{\right)}^{4}\phantom{\rule{0ex}{0ex}}=\frac{k\left(k+1\right)\left(2k+1\right)\left(3{k}^{2}+3k-1\right)}{30}+\left(k+1{\right)}^{4}\phantom{\rule{0ex}{0ex}}=\frac{\left(k+1\right)}{30}.\left[k\left(2k+1\right)\left(3{k}^{2}+3k-1\right)+30\left(k+1{\right)}^{3}\right]\phantom{\rule{0ex}{0ex}}=\frac{\left(k+1\right)}{30}.\left[6{k}^{4}+6{k}^{3}-2{k}^{2}+3{k}^{3}+3{k}^{2}-k+30.\left\{{k}^{3}+3{k}^{2}+3k+1\right\}\right]\phantom{\rule{0ex}{0ex}}=\frac{\left(k+1\right)}{30}.\left[6{k}^{4}+9{k}^{3}+{k}^{2}-k+30{k}^{3}+90{k}^{2}+90k+30\right]\phantom{\rule{0ex}{0ex}}=\frac{\left(k+1\right)}{30}.\left[6{k}^{4}+39{k}^{3}+91{k}^{2}+89k+30\right]$
now we will find the factor of $g\left(k\right)=6{k}^{4}+39{k}^{3}+91{k}^{2}+89k+30$
$g\left(-2\right)=6*\left(-2{\right)}^{4}+39*\left(-2{\right)}^{3}+91*\left(-2{\right)}^{2}+89*\left(-2\right)+30\phantom{\rule{0ex}{0ex}}=96-312+364-178+30\phantom{\rule{0ex}{0ex}}=490-490\phantom{\rule{0ex}{0ex}}=0$
thus $\left(k+2\right)$ is a factor of g(k).

now k = -3/2 is a factor of g(k).
thus by the factorization we can write:
$g\left(k\right)=\left(k+2\right)\left(2k+3\right)\left(3{k}^{2}+9k+5\right)$
therefore
${1}^{4}+{2}^{4}+{3}^{4}+...........+{k}^{4}+\left(k+1{\right)}^{4}\phantom{\rule{0ex}{0ex}}=\frac{\left(k+1\right)}{30}.\left(k+2\right)\left(2k+3\right)\left(3{k}^{2}+9k+5\right)\phantom{\rule{0ex}{0ex}}=\frac{\left(k+1\right)}{30}.\left\{\left(k+1\right)+1\right\}.\left\{2.\left(k+1\right)+1\right\}.\left\{3.\left(k+1{\right)}^{2}+3\left(k+1\right)-1\right\}$
thus the given equation also holds for n = k+1.
thus it is true for all integer values of n.

hope this helps you

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