Question

Can you find the sum of the series 1*2+2*3+3*4+4*5+5*6+----------+n(n+1)?

Answer 1

Step-by-step explanation:

Given:-

1*2+2*3+3*4+4*5+5*6+---+n(n+1)

To find:-

Find the sum of the series 1*2+2*3+3*4+4*5+5*6+---+n(n+1)?

Solution:-

Given series is 1×2+2×3+3×4+...+n(n+1)

Last term =n(n+1)=n²+n

Sum of first ( n²+n ) terms

=>Sum of squares of first 'n' natural numbers + Sum of first 'n' natural numbers

=>[n(n+1)(2n+1)/6]+[n(n+1)/2]

LCM of 6 and 2=6

=>[n(n+1)(2n+1)+3n(n+1)]/6

=>n(n+1)[2n+1+3]/6

=>n(n+1)(2n+4)/6

=>2n(n+1)(n+2)/6

=>n(n+1)(n+2)/3

Answer:-

1*2+2*3+3*4+---+n(n+1)=n(n+1)(n+2)/3