Dec 9, 2014 · 65 $\begingroup$ This question already has answers here: Proving the identity $\sum_ {k=1}^n {k^3} = \big (\sum_ {k=1}^n k\big)^2$ without induction (31 answers)

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?

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 …

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

Jun 28, 2014 · Find the sum of all positive integers $a=2^n3^m$ where $n$ and $m$ are non-negative integers, for which $a^6$ is not a divisor of $6^a$.

Jan 20, 2015 · Recursion tree for $T (n)=T (\frac n3)+T (\frac {2n} {3})+cn$ Shortest path will be most left one, because it operates on lowest value, and the most right one will be the longest …

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. …

Sep 15, 2020 · where the summation is over all triples (n1,n2,n3) (n 1, n 2, n 3) of non-negative integers with sum n n. This I know how to prove: Use the definition of Multinomial Theorem to …

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 …

Oct 30, 2012 · Let n3 be the number of 3-Sylow subgroups of G. then n3=1 or n3=4 if n3=1 we have 1 3-sylow group of order 9. and it is also a normal group (from sylow theorem ) if n2=1 …