Question

如何证明级数Sum[((k + 1)/(1 + p) - 1/2)*(k + 1)^p - (k/(1 + p) + 1

Answer 1

Sum[k + 1)/(1 + p) - 1/2)*(k + 1)^p - (k/(1 + p) + 1/2)*k^p

=Sum[((k + 1)^(p+1)/(1 + p) - (1/2)*(k + 1)^p -k^(p+1)/(1 + p) -( 1/2)*k^p}

=Sum[(k + 1)^(p+1)-k^(p+1)]/(1 + p)-0.5[(k + 1)^p+k^p]

=[n^(p+1)/(1 + p)-0.5SUM[(k + 1)^p+k^p]

=n^(p+1)/(1+p)-0.5*n^p-(1^p+2^p+……+（n-1)^p+n^p}

=(n+1)*n^p/(2+2p)--(1^p+2^p+……+（n-1)^p+n^p}