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  1. Proving $1^3+ 2^3 + \cdots + n^3 = \left (\frac {n (n+1)} …

    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) = …

  2. how to prove that $f (n)=n^3+n\log^2n$ = $\theta (n^3)$?

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

  3. Use mathematical induction to prove that $n^ 3 − n$ is divisible …

    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

  4. summation - Prove that $1^3 + 2^3 + ... + n^3 = (1+ 2

    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- …

  5. elementary number theory - Mathematics Stack Exchange

    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 …

  6. Big-O Notation - Prove that $n^2 + 2n + 3$ is $\mathcal O (n^2)$

    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 ...

  7. Prove that $ n^3 + 5n$ is divisible by 6 for all $n\in \textbf {N ...

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

  8. algorithms - How to arrange functions in increasing order of …

    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 ...

  9. calculus - Summation of $n/3^n$ - Mathematics Stack Exchange

    $$\\sum_{n=1}^\\infty \\frac{n}{3^n}$$ How do you find the sum? I don't know how to start this problem and no other website I found talks about a problem like this.

  10. calculus - Use the sum of the first ten terms to approximate the ...

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