# just when i thought everything was plain sailing....

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#1
these come up on my proof by induction worksheet .

had a stab at the first question lol then spent like another hour thinking....wtf!!!!

question two was ok, just that i mustve done something wrong somewhere because what i needed to prove and what i got are not the same lol (i subbed numbers in to see if it worked out before further rearrangement).

would be grateful for any help!!!!!! 0
15 years ago
#2
2.
Assume it's true for n=k:
sum[f(r)] = f(1) + f(2) + ... + f(k) = 1/8 - [3^(k+1)]/(k+4)!
Let n=k+1:
sum[f(r)] = f(1) + f(2) + ... + f(k) + f(k+1) = {f(1) + f(2) + ... + f(k)} + f(k+1)
= 1/8 - [3^(k+1)]/(k+4)! + f(k+1)
= 1/8 - [3^(k+1)]/(k+4)! + [3^(k+1)(k+2)]/(k+5)! [and (k+5)!=(k+5)(k+4)!]
= 1/8 - [3^(k+1)](k+5)/(k+5)(k+4)! + [3^(k+1)(k+2)]/(k+5)(k+4)!
= 1/8 + {3^(k+1)(k+2)-3^(k+1)(k+5)}/(k+5)!
= 1/8 + 3^(k+1)[(k+2)-(k+5)/(k+5)!
= 1/8 + 3^(k+1)(-3)/(k+5)!
= 1/8 - 3^(k+1)(3)/(k+5)!
= 1/8 - 3^(k+2)/(k+5)!
= 1/8 - 3^[(k+1)+1]/[(k+1)+4]!
=> It's true for n=k+1 if it's true for n=k.
Now let n=1:
3(2)/5! = 6/120 = 1/20
1/8 - 9/120 = 1/20
=> True for n=1, and thus, by induction, true for all positive integral n.

I'd do the first one too, but I'll leave it so you can practice. It's basically the same method as this one, but I believe you'll need to use some sort of identity. (And I don't feel like doing that! )
0
#3
*dawns on me* aaaaaaaaaaaahhhhhh double angle formulae?
0
#4
(Original post by dvs)
2.
Assume it's true for n=k:
sum[f(r)] = f(1) + f(2) + ... + f(k) = 1/8 - [3^(k+1)]/(k+4)!
Let n=k+1:
sum[f(r)] = f(1) + f(2) + ... + f(k) + f(k+1) = {f(1) + f(2) + ... + f(k)} + f(k+1)
= 1/8 - [3^(k+1)]/(k+4)! + f(k+1)
= 1/8 - [3^(k+1)]/(k+4)! + [3^(k+1)(k+2)]/(k+5)! [and (k+5)!=(k+5)(k+4)!]
= 1/8 - [3^(k+1)](k+5)/(k+5)(k+4)! + [3^(k+1)(k+2)]/(k+5)(k+4)!
= 1/8 + {3^(k+1)(k+2)-3^(k+1)(k+5)}/(k+5)!
= 1/8 + 3^(k+1)[(k+2)-(k+5)/(k+5)!
= 1/8 + 3^(k+1)(-3)/(k+5)!
= 1/8 - 3^(k+1)(3)/(k+5)!
= 1/8 - 3^(k+2)/(k+5)!
= 1/8 - 3^[(k+1)+1]/[(k+1)+4]!
=> It's true for n=k+1 if it's true for n=k.
Now let n=1:
3(2)/5! = 6/120 = 1/20
1/8 - 9/120 = 1/20
=> True for n=1, and thus, by induction, true for all positive integral n.

I'd do the first one too, but I'll leave it so you can practice. It's basically the same method as this one, but I believe you'll need to use some sort of identity. (And I don't feel like doing that! )
i see where i went wrong now grrrrrrrrr thanks a lot though!!!!!
0
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