Rasing (surds)^2000 without a calculator! (1 Viewer)

[Sharky]

Hey, i've got this question that's come from a friend. Can anyone lend me a hand? I've already worked out solutions to the first part, and a rough one for the second part, but the third still evades me.

Thanx.

NOTE: No calculators for any of these either!


i) If S=(Sqrt(50)+7)^2000+(Sqrt(50)-7)^2000, show that 'S' is even.

ii) Show that 0 < Sqrt(50)-7 < 1/10.

iii) Hence or otherwise, prove that the 1st 2000 digits after the decimal place of the expansion of (Sqrt(50)+7)^2000 are all 9's.

 

[wogboy]

since 0 < sqrt(50) - 7 < 1/10 (part ii)
0 < (sqrt(50) - 7)^2000 < 10^(-2000)

let S = (sqrt(50) + 7)^2000 + (sqrt(50) - 7)^2000
-> S - (sqrt(50) - 7)^2000 = sqrt(50) + 7
-> 0 < (sqrt(50) + 7)^2000 < 10^(-2000)
-> S - 10^(-2000) < S - (sqrt(50) + 7)^2000 < S
-> S - 10^(-2000) < (sqrt(50) + 7)^2000 < S

Suppose A is the set of all numbers greater than (S-1) and less than S, where the first 2000 digits after the decimal point are 9.

let x = (sqrt(50) + 7)^2000
-> S - 10^(-2000) < x < S
-> x > S - 10^(-2000) AND x < S

x is in the set A if:
x < S AND x >= (S - 1) + 9*sum{k from 1 to 2000} 10^(-k)
and we know from above that the first condition (x < S) is satisfied. So:
x is in the set A if:
x >= S - 1 + 9*sum{k from 1 to 2000} 10^(-k)

but we know x > S - 10^(-2000),
so x is in the set A if:
S - 10^(-2000) >= S - 1 + 9*sum{k from 1 to 2000} 10^(-k)
i.e. - 10^(-2000) >= - 1 + 9*sum{k from 1 to 2000} 10^(-k)
i.e. 1 - 10^(-2000) >= 9*sum{k from 1 to 2000} 10^(-k) (which is the sum of a geometric progression)
i.e. 1 - 10^(-2000) >= 0.9*(1 - 10^-(2000)) / (0.9)
i.e. 1 - 10^(-2000) >= 1 - 10^(-2000)
which is obviously true!!

so x is in the set A.
so the first 2000 digits of (sqrt(50) + 7)^2000 are 9.
(A bit of a messy proof, I'll leave it for you to clean up )

 

[Affinity]

(7 + 1/10)^2 = 49 + 1/100 + 14/10 > 50
therefore 2 is true