Without assuming the formula, find the sum of the series $1^{7} + 2^{7} + 3^{7} + . . . . + n^{7}$ .

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Solution

$\Rightarrow$ Assume
$1^{7} + 2^{7} + . . . . . + n^{7} = A_{0} n^{8} + A_{1} n^{7} + A_{2} n^{6} + . . . . + A_{7} n + A_{8};$
${(n + 1)}^{7} = A_{0} {{(n + 1)}^{8} - n^{8}} + A_{1} {{(n + 1)}^{7} - n^{7}} + . . . . + A_{7}$
$1 = 8 A_{0}; A_{0} = \frac{1}{8}$
$7 = 28 A_{0} + 7 A_{1}; A_{1} = \frac{1}{2}$
$21 = 56 A_{0} + 21 A_{1} + 6 A_{2}; A_{2} = \frac{7}{12}$
$35 = 70 A_{0} + 35 A_{1} + 15 A_{2} + 5 A_{3}; A_{3} = 0$
$35 = 56_{0} + 35 A_{1} + 20 A_{2} + 4 A_{4}; A_{4} = - \frac{7}{24}$
$21 = 28 A_{0} + 21 A_{1} + 15 A_{2} + 6 A_{4} + 3 A_{5}; A_{5} = 0$
$7 = 8 A_{0} + 7 A_{1} + 6 A_{2} + 4 A_{4} + 2 A_{8}; A_{6} = \frac{1}{12}$
$1 = A_{0} + A_{1} + A_{2} + A_{4} + A_{6} + A_{7}; A_{7} = 0$
Putting $n = 1$
$1 = A_{0} + A_{1} + A_{2} + A_{4} + A_{6} + A_{8}$
$A_{8} = 0$
$S_{n} = \frac{n^{8}}{8} + \frac{n^{7}}{2} + \frac{7 n^{6}}{12} - \frac{7 n^{4}}{24} + \frac{n^{2}}{12}$

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