17. Prove by induction that 1³+23++n³= (1+2+...+n)² for all positive integers n, that is, the sum of the cubes of the first n integers is equal to the square of the sum of the first n integers.
17. Prove by induction that 1³+23++n³= (1+2+...+n)² for all positive integers n, that is, the sum of the cubes of the first n integers is equal to the square of the sum of the first n integers.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![17. Prove by induction that
1³ +23+ + n³ = (1+2+...+n)²
for all positive integers n, that is, the sum of the cubes of the first n
integers is equal to the square of the sum of the first n integers.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F30761ad5-6d22-4ff4-adb6-4f166a7ab52a%2F7ee11fdc-0ac0-4f5b-839d-a81f711a16e6%2F9pd17wi.jpeg&w=3840&q=75)
Transcribed Image Text:17. Prove by induction that
1³ +23+ + n³ = (1+2+...+n)²
for all positive integers n, that is, the sum of the cubes of the first n
integers is equal to the square of the sum of the first n integers.
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I don't understand how (1+2+.....+k)^2=(k(k+1)/2))^2. Please explain. Thank you
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