here is the general method, lets say you have n terms instead of 10, you can always subst the value of n at the end.

 

The general term of the series is

=r(r+1)(r+2)

=r³+3r²+2r

 

Now substitute r as 1,2,3....n to get the sum of the series.

.: Sn = (1)³ + 3(1)² + 2(1) + (2)³ + 3(2)² + 2(2).....n terms

= (1³ + 2³ + 3³... n terms) + 3(1² + 2² + 3²....n terms) + 2(1 + 2 + 3... n terms)

 

= using sum of cubes of first n nat nos.= n²(n+1)²/4

sum of squares of first n =n(n+1)(2n + 1) / 6  

sum of first n nat numbers = n(n+1)/2

 you get the sum of series as

 

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

 

.: for n = 10

we get the sum as 4290