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...1 maybe > 1). 2. Assume it to be true for

**a**generic term**n**. 3. Show that if it works for**n**it...2/2]^2 = 1 hence true for**n**= 1 assume true for**n**: ∑**j**^3 = [**n**(**n**+1)/2]^2 for i = 1 to**n**for**n**+1 ∑**j**^3...1 Answers · Science & Mathematics · 11/10/2009

general term is

**a**[**j**] = (**n**+1-**j**)***j**; s(**n**) =Σ**a**[**j**], where**j**=1 until**n**; ♠check**n**=2...3+1)*(3+2)/6 =10; good! ♣ s(**n**+1) =Σ ((**n**+1) +1 –**j**)***j**= Σ ((**n**+1 -**j**+1)***j**{**j**=1 until**n**+1} = = Σ((**n**+1 –**j**)***j**+**j**...2 Answers · Science & Mathematics · 04/09/2008

... that, if you have

**a**matrix of the form**A**= [**a**_1...ca_i + db_i...**a**_**n**] where the**a**_**j**'s (and ca_i + db_i) stand for the columns of the matrix**A**and c...4 Answers · Science & Mathematics · 11/08/2008

1) (1,

**j**) is**a**member of S for each**j**from**N**==> cardinality of S is at least as big as the cardinality of**N**...1 Answers · Science & Mathematics · 21/07/2010

**n**(**j**(x)) = 5/((-6 + 20/(-2 + x)) (-2 + x)) = 5/(32 - 6 x)1 Answers · Science & Mathematics · 30/07/2015

...

**j**∈ Z, m=3j } and B = {**n**∈ Z ; ∃ k ∈ Z,**n**=7k} then**A****n**B = {m ∈ Z ; ∃**j**∈ Z, m = 21j } 2. P({4,5,6}) = { {4,5;6} , { 4,5 } , { 4,6 } , { 5,6...1 Answers · Science & Mathematics · 22/12/2013

...list of all

**a**[k] ^**j**will still be completely distinct from the list**n**[k] and therefore (**a**^**j**,**n**) = (**a**,**n**) = 12 Answers · Science & Mathematics · 31/01/2012

...proposition. Suppose this is true for

**n**= k. Given p |**a**_1a_2a_3...**a**_k**a**_(k+1) = (**a**_1a_2a_3...**a**_k) * (**a**...inductive hypothesis implies that p |**a**_**j**for some**j**= 1,2,...,k. I hope that helps...3 Answers · Science & Mathematics · 08/09/2009

∑(

**n**on top, i=1 bottom)**a**_i =**a**_1 +**a**_2 + ... +**a**_**n**but ∑(**n**top, i=1 bottom)**a**_**j**=**a**_**j**+**a**_**j**+ ... +**a**_**j**=**n*****a**_**j**, since**j**is different from the index variable i. I hope that helps!1 Answers · Science & Mathematics · 26/07/2009

...

**n**) has**a**subsequence (**n**_**j**) such that sin(**n**_**j**) → -1, which implies**a**_**n**_**j**=**n**_**j**^sin(**n**_**j**) → 0. We see**a**_**n**has**a**subsequence that goes...1 Answers · Science & Mathematics · 02/10/2008