, \quad \mathbf { b } = \left( \begin{array} { l }
1
Show mark scheme
Show mark scheme source
Question 2:
Answer Marks
Guidance
Answer/Working Marks
Guidance
\((n+1)^2-n^2=n^2+2n+1-n^2=2n+1\Rightarrow\) odd B1
[1]
\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+\cdots+\frac{2n+1}{n^2(n+1)^2}=\frac{2^2-1^2}{1^2.2^2}+\frac{3^2-2^2}{2^2.3^2}+\cdots+\frac{(n+1)^2-n^2}{n^2(n+1)^2}\) M1A1
\(=1-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+\frac{1}{3^2}-\frac{1}{4^2}+\cdots+\frac{1}{n^2}-\frac{1}{(n+1)^2}\) M1
\(=1-\frac{1}{(n+1)^2}\) A1
Sum to infinity \(=1\) A1\(\checkmark\)
[5]
Copy
## Question 2:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(n+1)^2-n^2=n^2+2n+1-n^2=2n+1\Rightarrow$ odd | B1 | [1] |
| $\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+\cdots+\frac{2n+1}{n^2(n+1)^2}=\frac{2^2-1^2}{1^2.2^2}+\frac{3^2-2^2}{2^2.3^2}+\cdots+\frac{(n+1)^2-n^2}{n^2(n+1)^2}$ | M1A1 | |
| $=1-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+\frac{1}{3^2}-\frac{1}{4^2}+\cdots+\frac{1}{n^2}-\frac{1}{(n+1)^2}$ | M1 | |
| $=1-\frac{1}{(n+1)^2}$ | A1 | |
| Sum to infinity $=1$ | A1$\checkmark$ | [5] |
---
Show LaTeX source
Copy
, \quad \mathbf { b } = \left( \begin{array} { l }
1 \\
1 \\
1
\end{array} \right) , \quad \mathbf { c } = \left( \begin{array} { r }
0 \\
1 \\
- 1
\end{array} \right) \quad \text { and } \quad \mathbf { d } = \left( \begin{array} { r }
2 \\
- 1 \\
1
\end{array} \right)
\hfill \mbox{\textit{CAIE FP1 2014 Q2 [6]}}