please open! i need major help! i have an exam tommorrow =(?

Question

1/2 + (1/2)^2 +(1/2)^3 +....+ (1/2)^K = 1-(1/2)^K

i have to prove that it works for K and K+1
i need help!
i can do the k but not the k+1

Answer 1

S(n) = [a(1-r^n )]/(1-r)
=0.5(1-0.5^k)/0.5
=1-(0.5)^k
T(k+1)=ar^(n-1)
=o.5(0.5)^k=0.5^(k+1)
S(k+1)=1-0.5^k+0.5^(k+1)
=1-[0.5^(k+1){2-1}]
=1-0.5^(k+1)
S(k+1)=[0.5(1-(0.5)^(k+1)]/0.5
=1-(0.5)^(k+1)
Both formulae derived are equal therefore this is true for k+1

Answer 2

This is proof by induction.
If P(k) is given proposition, have to prove that P(1) is true and P(k + 1) is true:
Consider P(1) ie k = 1
LHS = 1/2
RHS = 1 - 1/2 = 1/2
Thus P(1) is true

Consider P(k + 1):-
1/2 + (1/2)² + --(1/2)^(k + 1) = 1 - (1/2)^(k + 1)

LHS = (1/2) + (1/2)² + --(1/2)^k + (1/2)^(k + 1)
RHS = 1 - (1/2)^k + (1/2)^(k + 1)

RHS = 1 - (1/2)^k(1 - (1/2))
RHS = 1 - (1/2)^k x (1/2)
RHS = 1 - (1/2)^(k + 1)
Thus P(k + 1) is true
Thus P(1) is true and P(k + 1) is true.
Therefore P(k) is true

Answer 3

Using induction:

let p(X) = 1/2 + (1/2)^2 +(1/2)^3 +....+ (1/2)^X = 1-(1/2)^X

Step 1
put X=1
Thus, p(1)
1/2 = 1 - (1/2) ^ 1
which is true.

Step 2
put X=k
1/2 + (1/2)^2 +(1/2)^3 +....+ (1/2)^K = 1-(1/2)^K
We asume this to be true.

Step 3
put X=(K+1)
1/2 + (1/2)^2 +(1/2)^3 +....+ (1/2)^K + (1/2)^{K+1} = 1-(1/2)^{K+1}
now 1/2 + (1/2)^2 +(1/2)^3 +....+ (1/2)^K = 1-(1/2)^K [from step 2]

Thus,
1-(1/2)^K + (1/2)^{K+1} = 1-(1/2)^{K+1}
consider the LHS,
= 1 - 1/2^k + 1/2^{k+1}
= 1 - 1/2^k + 1/2^k.2 //note this step
= 1- 1/2^k [1 - 1/2 ] // taking -1/k^2 common
= 1 - 1/2^k [1/2]
=1 - 1/2^k.{2} //since x^(n).x = x^(n+1) [law of indices]
= 1 - 1/2^(k+1)
= rhs
hence proved.

Answer 4

what do you mean you have to prove it works for k and k+1?

k supposedly can range over any natural number, so what's the problem for k+1. are you trying to do an inductive proof?