If $u_0=0, u_ {k+1}=u_k+3^k$. Prove that $u_n=n3^ {n-1}$
Jan 30, 2026 · As pointed out by @naturallyInconsistent and @John Omielan the given recursion does not have $$ u_n = n3^ {n-1} $$ The given recursion $$ u_ {k+1}=u_k+3^k$$ gives $$ u_n = 3^ {n-1} …
$\\sum_{m=1}^{\\infty}\\sum_{n=1}^{\\infty} \\frac{m²n}{n3^m +m3^n}$
Sep 8, 2020 · $\sum_ {m=1}^ {\infty}\sum_ {n=1}^ {\infty} \frac {m²n} {n3^m +m3^n}$. I replaced m by n,n by m and sum both which gives term $\frac {mn (m+n)} {n3^m +m3^n}$.how to do further?
Prove that $1^3 + 2^3 + ... + n^3 = (1+ 2 + ... + n)^2$
HINT: You want that last expression to turn out to be $\big (1+2+\ldots+k+ (k+1)\big)^2$, so you want $ (k+1)^3$ to be equal to the difference $$\big (1+2+\ldots+k+ (k+1)\big)^2- (1+2+\ldots+k)^2\;.$$ …
How to prove $ \sum_ {k=3}^n \binom nk \binom k3 =\binom n3 2
$$ \sum_ {k=3}^n \binom nk \binom k3 = \binom n3 2^ {n-3} $$ It seems that some terms in the binomial coefficients cancel out: $$\binom nk \binom k3 = \frac {n!} {k! (n-k)!} \cdot \frac {k!} { (k-3)!3!} = \frac …
Prove that $2^n3^ {2n}-1$ is always divisible by 17
7 Prove that $2^n3^ {2n} -1$ is always divisible by $17$. I am very new to proofs and i was considering using proof by induction but I am not sure how to. I know you have to start by verifying the statement …
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 …
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}\
recreational mathematics - Mathematics Stack Exchange
Nov 30, 2023 · 1. It is unfortunately absolutely hopeless to determine the FIRST digits of Graham's number. 2. TREE (3) is a completely different league , unimaginably larger than Graham's number. 3. …
Recursive definition of natural numbers - Mathematics Stack Exchange
Jul 11, 2015 · I'm doing the exercises in Algorithms and Data Structures in Java, Second Edition, by Adam Drozdek. One question is: The set of natural numbers $\\mathbb{N}$ defined at the beginning …
Proof that $n^3+2n$ is divisible by $3$
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...