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Question 0510
Sigma Notation
2010 Paper 2 Question 2 Variant
Question
Generate new
(a)
Prove by the method of differences that
∑
r
=
4
n
1
(
r
−
2
)
r
=
5
12
−
1
2
n
−
2
−
1
2
n
.
\sum_{r=4}^n \frac{1}{(r - 2) r} = \frac{5}{12} - \frac{ 1 }{ 2 n - 2 } - \frac{ 1 }{ 2 n }.
r
=
4
∑
n
(
r
−
2
)
r
1
=
12
5
−
2
n
−
2
1
−
2
n
1
.
[4]
(b)
Explain why
∑
r
=
4
∞
1
(
r
−
2
)
r
{\displaystyle \sum_{r=4}^\infty \frac{1}{(r - 2) r}}
r
=
4
∑
∞
(
r
−
2
)
r
1
is a convergent series, and state the value of the sum to infinity.
[2]
Answer
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