The value of the sum 1 2 3+2 3 4+3 4 5+ upto n terms is equal to

Step-by-step solution

Let given series be S=1 2 3+2 3 4+3 4 5+ n terms Now, general term aligned & T_n= 1+(n-1) 1 2+(n-1) 1 3+(n-1) 1 & =(1+n-1)(2+n-1)(3+n-1) & =n(n+1)(n+2) & =n (n^2+3 n+2 ) & =n^3+3 n^2+2 n aligned Now, aligned & Sum of the series S = T_n= (n^3+3 n^2+2 n ) & = n^3+3 n^2+2 n & = n^2(n+1)^2 4 + 3 n(n+1)(2 n+1) 6 + 2 n(n+1) 2 & = n(n+1) 2 [ n(n+1) 2 +(2 n+1)+2 ] aligned aligned & = n(n+1) 2 [ n^2+n+4 n+2+4 2 ] & = n(n+1) (n^2+5 n+6 ) 4 & S = n(n+1)(n+2)(n+3) 4 aligned