Question
1*2+2*3+3*4+4*5+…+n*(n+1)=?
Let S = 1*2+2*3+3*4+4*5+…+n*(n+1) … ( 1)
Use.
1 + 2 + … + n = n*( n + 1 ) / 2
and
1^2 + 2^2 + … + n^2 = [ n*(n+1)*(2n+1)/6 ]
So
S = Sum [ k =1 to k =n for k *( k + 1 ) ]
= Sum [ k =1 to k =n for k^2 ] + Sum [ k = 1 to k = n for k ]
= [1^2 + 2^2 + … + n^2 ] + [ 1 + 2 + … + n ]
= [ n*(n+1)*(2n+1)/6 ] + [ n*( n + 1 ) / 2 ] by employing the “formulas above ]
= [ n*(n + 1/2 ) ] * [ (2n + 1) / 3 + 1 ]
= [ n*(n + 1) /2 ] * [ (2n + 4) / 3 ]
= [ n*(n + 1) ] * [ (n + 2)*2 / 6 ]
= [ n*(n + 1 )*(n + 2) ] / 3
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