Let a_1 = 1$a_1 = 1$ and for n geq 1$n \geq 1$ , a_n+1 = frac12 a_n + fracn^2 - 2n - 1n^2 (n + 1)^2$a_{n+1} = \frac{1}{2} a_n + \frac{n^2 - 2n - 1}{n^2 (n + 1)^2}$ . Then left|sum_n=1^inftyleft(a_n - frac2n^2right)right|$\left|\sum_{n=1}^{\infty}\left(a_n - \frac{2}{n^2}\right)\right|$ is equal

### Related Formula Partial fraction decomposition for telescopic summing: frac2n^2 - (n+1)^2 + 1 dotsdots text structures directly cancel in series expansions.$$\frac{2n^2 - (n+1)^2 + 1 \dots}{\dots} \text{ structures directly cancel in series expansions.}$$ ### Core Logic Given recurrence: a_n+1 - frac12a_n = fracn^2 - 2n - 1n^2(n+1)^2$$a_{n+1} - \frac{1}{2}a_n = \frac{n^2 - 2n - 1}{n^2(n+1)^2}$$ Rewrite the numerator to split the fraction: n^2 - 2n - 1 = 2n^2 - (n^2 + 2n + 1) = 2n^2 - (n+1)^2$$n^2 - 2n - 1 = 2n^2 - (n^2 + 2n + 1) = 2n^2 - (n+1)^2$$ a_n+1 - frac12a_n = frac2n^2 - (n+1)^2n^2(n+1)^2 = frac2(n+1)^2 - frac1n^2$$a_{n+1} - \frac{1}{2}a_n = \frac{2n^2 - (n+1)^2}{n^2(n+1)^2} = \frac{2}{(n+1)^2} - \frac{1}{n^2}$$ ### Step 1: Telescope generation Multiply both sides by appropriate powers of 2 to create a cancelling chain: For n=1$n=1$: a_2 - frac12a_1 = frac22^2 - frac11^2$a_2 - \frac{1}{2}a_1 = \frac{2}{2^2} - \frac{1}{1^2}$ For n=2$n=2$: multiply by 2 Rightarrow 2left[a_3 - frac12a_2 = frac23^2 - frac12^2right] Rightarrow 2a_3 - a_2 = frac2 times 23^2 - frac22^2$\Rightarrow 2\left[a_3 - \frac{1}{2}a_2 = \frac{2}{3^2} - \frac{1}{2^2}\right] \Rightarrow 2a_3 - a_2 = \frac{2 \times 2}{3^2} - \frac{2}{2^2}$ Wait, let's look at a cleaner telescopic scaling: a_n+1 - frac2(n+1)^2 = frac12 left(a_n - frac2n^2right)$a_{n+1} - \frac{2}{(n+1)^2} = \frac{1}{2} \left(a_n - \frac{2}{n^2}\right)$. ### Step 2: Identify Geometric Progression Let V_n = a_n - frac2n^2$V_n = a_n - \frac{2}{n^2}$. The recurrence gives V_n+1 = frac12 V_n$V_{n+1} = \frac{1}{2} V_n$. This proves V_n$V_n$ is a geometric progression with common ratio r = 1/2$r = 1/2$. First term V_1 = a_1 - frac21^2 = 1 - 2 = -1$V_1 = a_1 - \frac{2}{1^2} = 1 - 2 = -1$. ### Step 3: Infinite Summation We need left| sum_n=1^infty left( a_n - frac2n^2 right) right| = left| sum_n=1^infty V_n right|$\left| \sum_{n=1}^{\infty} \left( a_n - \frac{2}{n^2} \right) \right| = \left| \sum_{n=1}^{\infty} V_n \right|$. Since V_n$V_n$ is an infinite GP: S_infty = fracV_11 - r = frac-11 - 1/2 = frac-11/2 = -2$$S_{\infty} = \frac{V_1}{1 - r} = \frac{-1}{1 - 1/2} = \frac{-1}{1/2} = -2$$ Taking absolute value: |-2| = 2$|-2| = 2$ ### Pattern Recognition When dealing with rational fraction recurrences A_n+1 - k A_n = f(n) - k f(n-1)$A_{n+1} - k A_n = f(n) - k f(n-1)$, immediately substitute V_n = A_n - f(n)$V_n = A_n - f(n)$. This substitution instantly isolates a classical Geometric Progression. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Sequences and Series