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Elementary Algebra - Part 1
Almaz Khusnutdinov
•
Math enthusiast, love coding and mathematics
Published 10/16/2025
Telescoping Methods
Problem 1.1
Evaluate
S
n
=
∑
k
=
1
n
1
k
(
k
+
1
)
by telescoping.
Detailed
explanation:
1
k
(
k
+
1
)
=
1
k
−
1
k
+
1
.
⇒
S
n
=
(
1
−
1
2
)
+
(
1
2
−
1
3
)
+
⋯
+
(
1
n
−
1
n
+
1
)
=
1
−
1
n
+
1
.
Answer:
S
n
=
1
−
1
n
+
1
.
\text{Evaluate }S_n=\displaystyle\sum_{k=1}^{n}\frac{1}{k(k+1)}\ \text{by telescoping.}\\[4pt] \textbf{Detailed explanation: } \frac{1}{k(k+1)}=\frac{1}{k}-\frac{1}{k+1}.\\ \Rightarrow\ S_n=\Big(1-\frac12\Big)+\Big(\frac12-\frac13\Big)+\cdots+\Big(\frac{1}{n}-\frac{1}{n+1}\Big)=1-\frac{1}{n+1}.\\[4pt] \textbf{Answer: }S_n=1-\dfrac{1}{n+1}.
Evaluate
S
n
=
k
=
1
∑
n
k
(
k
+
1
)
1
by telescoping.
Detailed explanation:
k
(
k
+
1
)
1
=
k
1
−
k
+
1
1
.
⇒
S
n
=
(
1
−
2
1
)
+
(
2
1
−
3
1
)
+
⋯
+
(
n
1
−
n
+
1
1
)
=
1
−
n
+
1
1
.
Answer:
S
n
=
1
−
n
+
1
1
.
Problem 1.2
Evaluate
∑
k
=
1
n
1
k
(
k
+
1
)
(
k
+
2
)
.
Answer:
1
2
(
1
1
⋅
2
−
1
(
n
+
1
)
(
n
+
2
)
)
=
1
4
−
1
2
(
n
+
1
)
(
n
+
2
)
.
\text{Evaluate } \displaystyle\sum_{k=1}^{n}\frac{1}{k(k+1)(k+2)}.\\[4pt] \textbf{Answer: }\displaystyle \frac{1}{2}\!\left(\frac{1}{1\cdot2}-\frac{1}{(n+1)(n+2)}\right)=\frac{1}{4}-\frac{1}{2(n+1)(n+2)}.
Evaluate
k
=
1
∑
n
k
(
k
+
1
)
(
k
+
2
)
1
.
Answer:
2
1
(
1
⋅
2
1
−
(
n
+
1
)
(
n
+
2
)
1
)
=
4
1
−
2
(
n
+
1
)
(
n
+
2
)
1
.
Problem 1.3
Evaluate
∑
k
=
1
n
(
k
+
1
−
k
)
.
Answer:
n
+
1
−
1.
\text{Evaluate } \displaystyle\sum_{k=1}^{n}\big(\sqrt{k+1}-\sqrt{k}\big).\\[4pt] \textbf{Answer: }\displaystyle \sqrt{n+1}-1.
Evaluate
k
=
1
∑
n
(
k
+
1
−
k
)
.
Answer:
n
+
1
−
1.
Problem 1.4
Evaluate
∑
k
=
1
n
ln
(
k
+
1
k
)
.
Answer:
ln
(
n
+
1
)
.
\text{Evaluate } \displaystyle\sum_{k=1}^{n}\ln\!\left(\frac{k+1}{k}\right).\\[4pt] \textbf{Answer: }\displaystyle \ln(n+1).
Evaluate
k
=
1
∑
n
ln
(
k
k
+
1
)
.
Answer:
ln
(
n
+
1
)
.
Problem 1.5
Evaluate the product
∏
k
=
2
n
(
1
−
1
k
2
)
.
Answer:
n
+
1
2
n
.
\text{Evaluate the product } \displaystyle\prod_{k=2}^{n}\left(1-\frac{1}{k^{2}}\right).\\[4pt] \textbf{Answer: }\displaystyle \frac{n+1}{2n}.
Evaluate the product
k
=
2
∏
n
(
1
−
k
2
1
)
.
Answer:
2
n
n
+
1
.
Problem 1.6
Evaluate the product
∏
k
=
1
n
k
+
2
k
.
Answer:
(
n
+
2
)
!
/
2
!
n
!
=
(
n
+
1
)
(
n
+
2
)
2
.
\text{Evaluate the product } \displaystyle\prod_{k=1}^{n}\frac{k+2}{k}.\\[4pt] \textbf{Answer: }\displaystyle \frac{(n+2)!/2!}{n!}=\frac{(n+1)(n+2)}{2}.
Evaluate the product
k
=
1
∏
n
k
k
+
2
.
Answer:
n
!
(
n
+
2
)!
/2
!
=
2
(
n
+
1
)
(
n
+
2
)
.
Problem 1.7
Evaluate the product
∏
k
=
1
n
k
+
1
k
+
3
.
Answer:
2
⋅
3
⋯
(
n
+
1
)
4
⋅
5
⋯
(
n
+
3
)
=
3
!
(
n
+
3
)
(
n
+
2
)
=
6
(
n
+
2
)
(
n
+
3
)
.
\text{Evaluate the product } \displaystyle\prod_{k=1}^{n}\frac{k+1}{k+3}.\\[4pt] \textbf{Answer: }\displaystyle \frac{2\cdot3\cdots(n+1)}{4\cdot5\cdots(n+3)}=\frac{3!}{(n+3)(n+2)}=\frac{6}{(n+2)(n+3)}.
Evaluate the product
k
=
1
∏
n
k
+
3
k
+
1
.
Answer:
4
⋅
5
⋯
(
n
+
3
)
2
⋅
3
⋯
(
n
+
1
)
=
(
n
+
3
)
(
n
+
2
)
3
!
=
(
n
+
2
)
(
n
+
3
)
6
.
Problem 1.8
Evaluate
∑
k
=
1
n
1
k
(
k
+
2
)
.
Answer:
1
2
∑
k
=
1
n
(
1
k
−
1
k
+
2
)
=
3
4
−
1
2
(
1
n
+
1
+
1
n
+
2
)
.
\text{Evaluate } \displaystyle\sum_{k=1}^{n}\frac{1}{k(k+2)}.\\[4pt] \textbf{Answer: }\displaystyle \frac12\sum_{k=1}^{n}\!\left(\frac{1}{k}-\frac{1}{k+2}\right)=\frac{3}{4}-\frac{1}{2}\!\left(\frac{1}{n+1}+\frac{1}{n+2}\right).
Evaluate
k
=
1
∑
n
k
(
k
+
2
)
1
.
Answer:
2
1
k
=
1
∑
n
(
k
1
−
k
+
2
1
)
=
4
3
−
2
1
(
n
+
1
1
+
n
+
2
1
)
.
Table of Contents
Telescoping Methods
Problem 1.1
Problem 1.2
Problem 1.3
Problem 1.4
Problem 1.5
Problem 1.6
Problem 1.7
Problem 1.8
Elementary Algebra - Part 1 - nerjik.com
