<p>Given that&nbsp;${{u}_{n}}=\frac{{{2}^{n}}}{n}$, show that&nbsp;${{u}_{n+1}}-{{u}_{n}}=\frac{{{2}^{n}}\left( n-1 \right)}{n\left( n+1 \right)}$ where&nbsp;$n$ is a positive integer.</p>
<p>Hence, show that&nbsp;$\sum\limits_{n=1}^{N}{\frac{{{2}^{n}}\left( n-1 \right)}{n\left( n+1 \right)}}=\frac{{{2}^{N+1}}}{N+1}-2$.</p>

Given that ${{u}_{n}}=\frac{{{2}^{n}}}{n}$, show that ${{u}_{n+1}}-{{u}_{n}}=\frac{{{2}^{n}}\left( n-1 \right)}{n\left( n+1 \right)}$ where $n$ is a positive integer.

Hence, show that $\sum\limits_{n=1}^{N}{\frac{{{2}^{n}}\left( n-1 \right)}{n\left( n+1 \right)}}=\frac{{{2}^{N+1}}}{N+1}-2$.

Method of Difference involving uₙ−uₙ₋₁.

Recurrence Relations and Summation

International Baccalaureate Higher Level Mathematics

IB HL Math

N1991 P1 Q3 (a)

$u_{n+1} - u_n = \frac{2^{n+1}}{n+1} - \frac{2^n}{n}$\n$= \frac{2^n \cdot 2(n) - 2^n(n+1)}{n(n+1)}$\n$= \frac{2^n(2n - (n+1))}{n(n+1)}$\n$= \frac{2^n(n-1)}{n(n+1)}$

$\sum_{n=1}^{N} \frac{2^n(n-1)}{n(n+1)} = \sum_{n=1}^{N}(u_{n+1} - u_n)$

Using method of differences (telescoping):\n$= (u_2 - u_1) + (u_3 - u_2) + (u_4 - u_3) + \ldots + (u_{N+1} - u_N)$\n$= u_{N+1} - u_1$

$= \frac{2^{N+1}}{N+1} - \frac{2^1}{1}$\n$= \boxed{\frac{2^{N+1}}{N+1} - 2}$

N1991 P1 Q3 (a)

N1991 P1 Q3 (a)

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