Proving $1^3+ 2^3 + \\cdots + n^3
Dec 9, 2014 · The result now follows immediately by F(n) = (n(n + 1)/2)2 ⇒ F(n) − F(n − 1) = n3 F (n) = (n (n + 1) / 2) 2 ⇒ F (n) F (n 1) = n 3 The theorem reduces the proof to a trivial …
induction - Prove that $ n^3 + 5n$ is divisible by 6 for all $n\in ...
So n=3k+1, the remainder of n3 n 3 when divided by 3 is 1 and for 5n 5 n it will be 2 2 and 2 + 1 = 3 2 + 1 = 3. And for n = 3k + 2 n = 3 k + 2, the n3 n 3 leaves remainder 2 when divided by 3 …
show ${n\\brace n-2} = \\binom n3 + 3\\binom n4$ combinatorially
Dec 20, 2025 · Use combinatorial reasoning to show $$ {n\brace n-2} = \binom n3 + 3\binom n4$$ where the Stirling number is the number of partitions of $ [n]$ into $n-2$ parts.
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
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- …
Series convergence test, $\sum_ {n=1}^ {\infty} \frac { (x-2)^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_{m=1}^{\\infty}\\sum_{n=1}^{\\infty} \\frac{m²n}{n3^m …
Sep 8, 2020 · $\sum_ {m=1}^ {\infty}\sum_ {n=1}^ {\infty} \frac {m²n} {n3^m +m3^n}$. I replaced m by n,n by m and sum both which gives term $\frac {mn (m+n)} {n3^m +m3^n}$.how to do further?
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 ...
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 …
For how many positive values of $n$ are both $\frac n3$ and $3n
Oct 29, 2017 · 112 values is the number of positive values whose n/3 and n*3 both are 4-digit numbers.