Question 169946
taking the formula n+1,n+3,n+5....where n=1,2,3,4,........to get even integers we have 
2,4,6,8,.......48 this becomes an arithmetic series ie
2+4+6+8+......+48 where first term(a)=2,common diffrence(d)=2 
we first find the number of terms in this seies
using the formula {{{L=(a+(n-1)d)}}} where L is the last term.
48=2+(n-1)2
n=24 therefore there are 24 terms.
Now we find the the sum using the formula 

{{{S=(n/2)(2a+(n-1)d))}}}where S=sum n=number of terms
S=[24/2(2*2+(24-1)2]
 =12(4+46)
 =600
therefore the sum is 600