Sequence (1/(2*3))+(1/(3*4))+(1/(4*5))+…+(1/((n+1)*(n+2))) The sum of the first 60 items of +... is ((a)/(b)), where a and b are relatively prime positive integers, then a+b=how much.
数列(1/(2*3))+(1/(3*4))+(1/(4*5))+…+(1/((n+1)*(n+2)))+…的前60项的和是((a)/(b))，其中a和b是互质的正整数，那么a+b=多少．
<gadget id="calculator">15 + 31</gadget>
<output>46</output>

<result>46</result>