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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
=
1
n
1
(
r
+
3
)
(
r
+
5
)
=
9
40
−
1
2
n
+
8
−
1
2
n
+
10
.
\sum_{r=1}^n \frac{1}{(r + 3) (r + 5)} = \frac{9}{40} - \frac{ 1 }{ 2 n + 8 } - \frac{ 1 }{ 2 n + 10 }.
r
=
1
∑
n
(
r
+
3
)
(
r
+
5
)
1
=
40
9
−
2
n
+
8
1
−
2
n
+
10
1
.
[4]
(b)
Explain why
∑
r
=
1
∞
1
(
r
+
3
)
(
r
+
5
)
{\displaystyle \sum_{r=1}^\infty \frac{1}{(r + 3) (r + 5)}}
r
=
1
∑
∞
(
r
+
3
)
(
r
+
5
)
1
is a convergent series, and state the value of the sum to infinity.
[2]
Answer
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