Question

Find the sum of 1.2.3+2.3.4+3.4.5+ ……….up to n terms ?

Answer 1

Here, A n for the equation is n(n+1)(n+2)

=(n^2+n)(n+2)

=n^3+3n^2+2n

Therefore, A 1 = 1^3+(3*1^2)+(2*1)

A 2 =2^3+(3*2^2)+(2*2)…

A n =n^3+(3*n^2)+(2*n)

So,

S n

=(1^3+2^3+3^3…+n^3)

+3(1^2+2^2+3^2…+n^2)+2(1+2+3+…+n)

={「n(n+1)」/2}^2

+3{「n(n+1)(2n+1)」/6}+2{「n(n+1)/2}

=[n(n+1)]/2{[n(n+1)]/2 + [3(2n+1)]/3+2}

=[n(n+1)]/2{[n^2+n+4n+2+4]/2}

=[n(n+1)]/4{n^2+5n+6}

=[n(n+1)]/4{n^2+3n+2n+6}

=[n(n+1)(n+2)(n+3)]/4