• :00Days
• :00Hours
• :00Mins
• 00Seconds
A new era for learning is coming soon

Select your language

Suggested languages for you:

Americas

Europe

Q17E

Expert-verified
Found in: Page 330

Discrete Mathematics and its Applications

Book edition 7th
Author(s) Kenneth H. Rosen
Pages 808 pages
ISBN 9780073383095

Answers without the blur.

Just sign up for free and you're in.

Prove that $\sum _{j=1}^{n} {j}^{4}=n\left(n+1\right)\left(2n+1\right)\left(3{n}^{2}+3n-1\right)/30$ whenever n is a positive integer.

$\sum _{j=1}^{n} {j}^{4}=n\left(n+1\right)\left(2n+1\right)\left(3{n}^{2}+3n-1\right)/30$

See the step by step solution

Step: 1

Let P(n) be the statement that ${1}^{4}+{2}^{4}+{3}^{4}+\dots +{n}^{4}=n\left(n+1\right)\left(2n+1\right)\left(3{n}^{2}+3n-1\right)/30$

P (1) is true because 1.2.3.5/30=1.

Step: 2

Assume that P (k) is true.

Then

$\begin{array}{r}{1}^{4}+{2}^{4}+{3}^{4}+\dots +{k}^{4}+\left(k+1{\right)}^{4}\\ =k\left(k+1\right)\left(2k+1\right)\left(3{k}^{2}+3k-1\right)/30+\left(k+1{\right)}^{4}\\ =\left[\left(k+1\right)/30\right]\left[k\left(2k+1\right)\left(3{k}^{2}+3k-1\right)+30\left(k+1{\right)}^{3}\right]\\ =\left[\left(k+1\right)/30\right]\left(6{k}^{4}+39{k}^{3}+91{k}^{2}+89k+30\right)\\ =\left[\left(k+1\right)/30\right]\left(k+2\right)\left(2k+3\right)\left[3\left(k+1{\right)}^{2}+3\left(k+1\right)-1\right]\end{array}$

This demonstrates that P(k+1) is true.

Want to see more solutions like these?

Sign up for free to discover our expert answers

Recommended explanations on Math Textbooks

94% of StudySmarter users get better grades.