Question 3
If $$\sum_{k=1}^{N} \frac{2k+1}{(k^2+k)^2} = 0.9999$$ then determine the value of $$N$$.
Correct Answer: 99
Solution
Since $$(k^2+k)^2 = k^2(k+1)^2$$ and $$2k+1 = (k+1)^2 - k^2$$, each term telescopes as $$\frac{1}{k^2} - \frac{1}{(k+1)^2}$$. The sum therefore equals $$1 - \frac{1}{(N+1)^2}$$, so $$\frac{1}{(N+1)^2} = 0.0001$$. This gives $$(N+1)^2 = 10000$$ and $$N = 99$$.
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