Oct 2, 2012 · I'm trying to sum the following series? $n(1 + n + n^2 + n^3 + n^4 +.......n^{n-1})$ Do you have any ideas?

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 …

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 …

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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+ …

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 …

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

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

Jun 5, 2021 · Take $2^ { [\frac n2] + [\frac n4] + ..... + [\frac n {2^k}]}$ where $2^k \le n < 2^ {k+1}$. Then multiple by $3^ { [\frac n3]+ [\frac n9] + .... + [\frac n {3^j}]}$ where $3^j\le n < 3^ …