Dec 9, 2014 · Hint $ $ First trivially inductively prove the Fundamental Theorem of Difference Calculus $$\rm\ F (n) = \sum_ {k\, =\, 1}^n f (k)\, \iff\, F (n) - F (n\!-\!1)\, =\, f (n),\ \ \, F (0) = 0\qquad$$ The …

HINT: You want that last expression to turn out to be $\big (1+2+\ldots+k+ (k+1)\big)^2$, so you want $ (k+1)^3$ to be equal to the difference $$\big (1+2+\ldots+k+ (k+1)\big)^2- (1+2+\ldots+k)^2\;.$$ …

Nov 21, 2018 · i didn't really understand the hint .. is this a way toprove that nlog2n/n3 <= 1 ?

Let n^3+2n = P (n). We know that P (0) is divisible by 3. The inductive step shows that P (n+1) = P (n) + (something divisible by 3). So if P (0) is divisible by 3, then P (1) is divisible by 3, and then...

Use mathematical induction to prove that n3 − n n 3 n is divisible by 3 whenever n is a positive integer. Ask Question Asked 9 years, 7 months ago Modified 7 years, 7 months ago

Jul 5, 2013 · I'm taking a course in Discrete Mathematics this summer, and my book doesn't offer a very good explanation of Big-O notation. I understand that if $f(x)$ is ...

Prove that $ n^3 + 5n $ is divisible by 6 for all $ n \in \textbf {N} $. I provide my proof below.

Given the following functions i need to arrange them in increasing order of growth a) $2^ {2^n}$ b) $2^ {n^2}$ c) $n^2 \log n$ d) $n$ e) $n^ {2^n}$ My first attempt ...

Oct 5, 2020 · By the way: The value of the sum is $12-24\log {3\over2}=2.268837405$.

Oct 29, 2017 · 112 values is the number of positive values whose n/3 and n*3 both are 4-digit numbers.