summation - Prove that $1^3 + 2^3 + ... + n^3 = (1+ 2 + ... + n)^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 $$\big (1+2+\ldots+k+ (k+1)\big)^2- (1+2+\ldots+k)^2\;.$$ …
Proving $1^3+ 2^3 + \cdots + n^3 = \left (\frac {n (n+1)} {2}\right)^2 ...
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 …
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 ?
Proving by induction that $1^3 + 2^3 + 3^3 + \ldots + n^3 = \left ...
Mar 25, 2013 · Need guidance on this proof by mathematical induction. I am new to this type of math and don't know how exactly to get started. $$ 1^3 + 2^3 + 3^3 + \ldots + n^3 = \left [\frac {n (n+1)} {2}\
Prove that $3^n$ is not $O(2^n)$ - Mathematics Stack Exchange
For any specific value of $n$, $O (2^n) = O (1) = O (3^n)$. The point of big-oh notation is that it says something informative when $n$ is a variable ranging over ...
Prove by induction that $1+2^n<3^n$ for $n \\ge 2$
Oct 5, 2017 · One of the problems in you proof is that you don't make use of the inductive hypothesis, at least not explicitly. The other problem is in the line: $$1 + 2\cdot 2^n = 2\cdot 3^n$$ This isn't true. In …
Proof that $n^3+2n$ is divisible by $3$ - Mathematics Stack Exchange
Let n^3+2n = P (n). We know that P (0) is divisible by 3. The inductive step shows that P (n+1) = P (n) + (something divisible by 3). So if P (0) is divisible by 3, then P (1) is divisible by 3, and then...
Use mathematical induction to prove that $n^ 3 − n$ is divisible by 3 ...
Solution: Let $P(n)$ be the proposition “$n^3−n$ is divisible by $3$ whenever $n$ is a positive integer”. Basis Step:The statement $P(1)$ is true because $1^3 ...
Series convergence test, $\sum_ {n=1}^ {\infty} \frac { (x-2)^n} {n3^n}$
Jun 28, 2020 · By the ratio test, every x value between -1 and 5 would make the series converge. we just need to find out whether x=-1, 5 makes it converge. x=-1: The series will look like this. $$\sum_ …
algorithms - How to arrange functions in increasing order of growth ...
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 ...