Arthur D. answered 08/14/14
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1+2+3+4+5+...+27+28+29+30+31=???
add the first and the last
add the second and the next to last
add the third and the next to the next to the last and so on
1+31=32
2+30=32
3+29=32
4+28=32
5+27=32
there are 15 of these sums and one lone number in the middle, the number 16
1-15, 16, 17-31
15*32=480, 480+16=496
another solution...
the sum of an arithmetic sequence is given by the formula...
S(n)=(n/2)[2a+(n-1)d] where n=31, a=1(first term), and d=1(difference between terms)
S(n)=(31/2)[2*1+(31-1)1]
S(n)=(31/2)(2+30)
S(n)=(31/2)(32)=(31)(16)=496