Without assuming the formula, find the sum of the series $1^{6} + 2^{6} + 3^{6} + . . . . + n^{6}$ .

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Solution

$\Rightarrow 1$ Assume $1^{6} + 2^{6} + 3^{6} + . . . . . = A_{0} n^{7} + A_{1} n^{6} + A_{2} n^{5} + . . . . + A r$
${(n + 1)}^{6} = A_{0} {{(n + 1)}^{7} - n^{7}} + A_{1} {{(n + 1)}^{6} - n^{6}}$
$1 = 7 A_{0}; A_{0} = \frac{1}{7}$
$6 = 21 A_{6} + 6 A_{1}; A_{1} = \frac{1}{2}$
$15 = 35 A_{0} + 15 A_{1} + 5 A_{2}$
$A_{0} = \frac{1}{2}$
$20 = 35 A_{0} + 20 A_{1} + 10 A_{2} + 4 A_{3};$
$A_{3} = 0$
$15 = 21 A_{0} + 15 A_{1} + 10 A_{2} + 3 A_{4}; A_{4} = - \frac{1}{6}$
$6 = 7 A_{0} + 6 A_{1} + 5 A_{2} + 3 A_{4} + 2 A_{5}; A_{5} = 0$
$1 = A_{0} + A_{1} + A_{2} + A_{4} + A_{5}; A_{6} = \frac{1}{42}$
By putting $n = 1$ we have;-
$1 = A_{0} + A_{1} + A_{2} + A_{4} + A_{6} + A_{7}$
$A_{7} = 0$
$S_{n} = \frac{n^{7}}{7} + \frac{n^{6}}{2} + \frac{n^{5}}{2} - \frac{n^{3}}{6} + \frac{n}{42}$

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