web2.0rechner
 

CPhill

avatar
BenutzernameCPhill
Punkte130554
Membership
Stats
Fragen 56
Antworten 43469

 #1
avatar+130554 
+5

Here's the graph........https://www.desmos.com/calculator/brwx1waam2

 

f(x) = 4  when x = about 2.6...proof 1.7^2.6 =  about 3.973

We could go through an iteration process, but here's a better way (hey, they said we could take shortcuts, didn't they??)

x = log(4) / log(1.7) = 2.612552871704276....I'm not that familiar with significant figures, so I'll let you trim as many off this as you need

And f(x) = 0.5 when x = about -1.3.....proof 1.7^(-1.3) = about 0.5016

Again....instead of an iteration process, use this

x = log(0.5) / log(1.7) = -1.306276435852138...again I'll let you worry about the sig figs

I think that's most of it....

 

17.01.2015
 #1
avatar+130554 
+10

Let a_1, a_2, . . . , a_{10} be an arithmetic sequence. If a_1 + a_3 + a_5 + a_7 + a_9 = 17 and a_2 + a_4 + a_6 + a_8 + a_{10} = 15, then find a_1.

Subtracting the second sum from the first, we have

(a2 - a1) + (a4 - a3) + (a6 - a5) + (a8 - a7) + (a10 - a9) = -2   or

  d           +      d       +      d       +     d        +      d        =  -2   or

5d = -2     →   d = -2/5

Also, the sum of the arithmetic series is given by

S = (10/2)(a1 + a10) = 5(a1 + a10) = 32     divide both sides by 5

So

a1 + a10 = 32/5 →  a10 = 32/5 - a1  ( 1)

And the last term is given by

a10 = a1 + d(9)    (2)

Therefore, setting (1) = (2), we have

32/5 - a1 = a1 + d(9)     ... solve for a1

(32/5 -2a1) = d(9)

32/5 -2a1 = (-2/5)(9) = -18/5  

32/5 + 18/5 = 2a1

2a1 = 50/5

a1 = 5

And a10=  32/5 - a1 =  32/5 - 5 = 1.4

Proof

a1 + a3 + a5 + a7 + a9  = 5 + 4.2 + 3.4 + 2.6 + 1.8 = 17

And since each of the other 5 terms  is (2/5) less than it's preceding term.....

[17 - 5(2/5)]  = [17 - 2] = 15  = the sum of the terms with even subscripts

 

17.01.2015
 #2
avatar+130554 
+10

Very nice, geno.....!!!

I'm not as crafty as geno, so I had to "brute-force" this one to see if I could discover another possible pattern....!!!

(1+2) + (1+3) + (1+4) + (1+5) + (1+6) + (1+7)

         +  (2+3) + (2+4) + (2+5) + (2+6) + (2+7)

                       + (3+4) + (3+5) + (3+6) + (3+7)

                                    + (4+5) + (4+6) + (4+7)

                                                 + (5+6) + (5+7)

                                                              + (6+7) 

-------------------------------------------------------------------

(1+2)+(3+6)+(6+12)+(10+20)+(15+30)+(21+42) =

Notice that the sum of the first terms in the parentheses is just 56 = (8 x 7) = 1(8 x 7)

And notice that the sum of the second terms in the parentheses is just  112   = 2(8 x 7)

So we have 1(8 x 7)+ 2(8 x 7) = 3(8 x 7)= 8(7 x 3) = 8 x 21  =  (7+1)C(7,2) = (8)C(7,2)

Notice that the "pattern," is just (n+1)C(n,2)  ....where n = 7

BTW....this result can be extended to any two element subset sums of the first n positive integers.....for instance.....the sum of the two element subsets of the first 10 integers =(10+1)C(10,2)  = 11(45) = 495 ......

----------------------------------------------------------------------------------------------------

See the inductive proof of this below........

   

17.01.2015