Sum of 6th power of n consecutive natural number

\[\sum_{n=1}^{n}n^6=1^6+2^6+3^6+4^6+.......+n^6=\frac{n(2n+1)(n+1)(3n^4+6n^3-3n+1)}{42}\]

 

for n=2\[\sum_{n=1}^{2}n^6=1^6+2^6=65\]

for n=3 \[\sum_{n=1}^{3}n^6=1^6+2^6+3^6=794\]

for n=5\[\sum_{n=1}^{5}n^6=1^6+2^6+3^6+4^6+5^6=20515\]

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