Question 14
If $$\frac{1}{1 \times 3}+\frac{1}{2 \times 4}+\frac{1}{3 \times 5}+\cdots+\frac{1}{n(n+2)}=\frac{3553}{4830}$$, then $$n$$ is
Solution
Since $$\frac{1}{k(k+2)}=\frac{1}{2}\left(\frac{1}{k}-\frac{1}{k+2}\right)$$, the sum telescopes to $$\frac{3}{4}-\frac{2n+3}{2(n+1)(n+2)}$$. For $$n=68$$, this equals $$\frac{3}{4}-\frac{139}{9660}=\frac{3553}{4830}$$. Hence $$n=68$$.
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