This question was uploaded on 14/12/21 on social media accounts.
This question is from Singapore Math Olympiad 2019 Open Round 1 (Question 10)
Solution:
Given equation:
\[a_{n+1}-a_n = 4n+2\]
After substituting values of n:
\[a_2-a_1 = 6\]
\[a_3-a_2 = 10\]
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\[a_{n-1}-a_{n-2} = 4(n-2)+2\]
\[a_{n}-a_{n-1} = 4(n-1)+2\]
After adding these all equation:
\[a_{n}-a_1 = 4\sum_1^{n-1}k+2n\]
\[a_{n}-2 = 4\frac{(n-1)(n)}{2}+2n\]
\[\Rightarrow a_n = 2n^2\]
Now,
\[\frac1{a_n-2} = \frac1{2n^2-2} = \frac12\frac1{(n-1)(n+1)}\]
\[\Rightarrow \frac1{a_n-2}= \frac14\left[\frac1{(n-1)}-\frac1{(n+1)}\right]\]
Given sum:
\[S = \frac1{a_2-2} + \frac1{a_3-2} + \frac1{a_4-2} + ......\]
\[S = \frac14\left[\left(\frac11-\frac13\right)+\left(\frac12-\frac14\right)+\left(\frac13-\frac15\right)+......+\left(\frac1{\infty}-\frac1{\infty}\right)\right]\]
\[\Rightarrow S = \frac14\left[\frac11+\frac12\right] = \frac38\]
