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 $$\big (1+2+\ldots+k+ (k+1)\big)^2- (1+2+\ldots+k)^2\;.$$ …

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?

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}\

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 …

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Apr 18, 2015 · First, show that this is true for n=0: 03+ (0+1)3+ (0+2)3=9 Second, assume that this is true for n: n3+ (n+1)3+ (n+2)3=9k Third, prove that this is true for n+1: (n+1)3+ (n+2)3+ (n+3)3= …

Oct 29, 2017 · 112 values is the number of positive values whose n/3 and n*3 both are 4-digit numbers.

Sep 15, 2020 · Problem Statement: Prove that for any positive integer $n$, $$\sum_ {n_1+n_2+n_3 = n} (-1)^ {n_1} = 1$$ where the summation is over all triples $ (n1, n2, n3)$ of non-negative integers with …

Jan 26, 2019 · If I understand correctly, any a looks like $$a = 2^n3^n$$ So we could also say any b looks like $$b = 2^u3^v, u, v \in \mathbb {Z}$$ Then $$ab = 2^n3^m2^u3^v$$ $$ab= 2^ {n+u}3^ …

Aug 21, 2011 · To Prove : $n! > (n/e)^n$ The question seems easy but it ain't; anyone up for it ?