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- …

Jul 5, 2013 · Would this book be Kenneth H. Rosen's "Discrete Mathematics and its Applications" by any chance? If so, I found all of the material easily digestible until this section, where I feel …

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) = …

Solution: Let $P(n)$ be the proposition “$n^3−n$ is divisible by $3$ whenever $n$ is a positive integer”. Basis Step:The statement $P(1)$ is true because $1^3 ...

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 …

Jun 28, 2020 · By the ratio test, every x value between -1 and 5 would make the series converge. we just need to find out whether x=-1, 5 makes it converge. x=-1: The series will look like this. …

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 $12-24\log {3\over2}=2.268837405$.

Jan 8, 2023 · Hi! Since the sum is over (2^k)* (n) I don't understand how you could apply the arithmetic progression for 2^k and just multiply n with it without applying the sum to n. In my …