| 1²³+2²³ + ... + n²³² = n²(n+1)² ५

Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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ABC
26.1) Use induction to prove the following statements.
a.) 1²³ +₂²³ + ... + n²³² =
n²(n+1)²
4
b.) 1.2 +2.3+3.4 + ... + (n-1)n=-
0³-n
3
JT₁+T₂+-+Tn = n(n+1)(n+₂), where Th= n(n+1) is the nth triangular
number
2
d. For
natural number n, write
1
D
every
An
Bn
asa
fraction in lowest terms. Prove that the denominator Br
divides ni. (Although, there are other ways to prove this statement,
should give a proof by induction.
you
11
1 + 1/²+1/3+
+---+
S
Transcribed Image Text:26.1) Use induction to prove the following statements. a.) 1²³ +₂²³ + ... + n²³² = n²(n+1)² 4 b.) 1.2 +2.3+3.4 + ... + (n-1)n=- 0³-n 3 JT₁+T₂+-+Tn = n(n+1)(n+₂), where Th= n(n+1) is the nth triangular number 2 d. For natural number n, write 1 D every An Bn asa fraction in lowest terms. Prove that the denominator Br divides ni. (Although, there are other ways to prove this statement, should give a proof by induction. you 11 1 + 1/²+1/3+ +---+ S
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