- We say that S(n) = 3- 1/2^(n-1). Show that this formula by help of mathematical induction is true in case n=1 and if n = k+1 ;
- S(1) = 2 ; S(1) = 3- 1/2^(1-1) = 3 - 1 = 2
- n = k + 1. S(k+1) = 2+1/2+1/4 +..+1/2^k + 1/2^k-1= S(k) + 1/2^k
- S(k) = 3- 1/2^(k-1). Then S(k+1) = S(k) + 1/2^k = 3 - 1/2^(k-1)+1/2^k =
= 3 - 1/(2^k * 2^-1) + 1/2^k = 3 - 2/2^k + 1/2^k = 3 - 1/2^k;
5.If we show to the first formula and substitute n = k+1, S(k+1) = 3- 1/2^(k+1-1) =
= 3 - 1/2^k
6. Saw that if the formula is true for n=k, then it is true for n = k+ 1,
thus, the formula is true for any n;
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