+466 Show that replacing k with k + 1 in 1 - 1/2k gives an expression equivalent to 1 - 1/2k + 1/2k+1
+129850 Show that replacing k with k + 1 in 1 - 1/2^k gives an expression equivalent to 1 - 1/2^k + 1/2^(k+1)
1 - 1/ [2^(k + 1)] =
1 - 1 / [2^k * 2] =
1 - 1/2^k (1/2) =
1 + (-1/2)(1/2^k) = [notice that -1/2 = -1 + 1/2]
1 + [ -1 + 1/2] [ 1/2^k] =
1 - 1/2^k + (1/2)(1/2^k) =
1 - 1/2^k + 1 / [ 2^k *2] =
1 - 1/2^k + 1/ 2^(k + 1)
+118667 Hi Shades :)
Show that replacing k with k + 1 in 1 - 1/2k gives an expression equivalent to 1 - 1/2k + 1/2k+1
Going back to front it is easy enough.
\(1-\frac{1}{2^k}+ \frac{1}{2 ^{k+1} } \\ =1-\left[\frac{1}{2^k}- \frac{1}{2 ^{k+1} } \right]\\ =1-\left[\frac{2}{2^{k+1}}- \frac{1}{2 ^{k+1} } \right]\\ =1-\left[\frac{1}{2^{k+1} } \right]\\ =1-\frac{1}{2^{k+1} } \\\)
Now I want to work out how to do it in the other direction ://
+129850 Show that replacing k with k + 1 in 1 - 1/2^k gives an expression equivalent to 1 - 1/2^k + 1/2^(k+1)
1 - 1/ [2^(k + 1)] =
1 - 1 / [2^k * 2] =
1 - 1/2^k (1/2) =
1 + (-1/2)(1/2^k) = [notice that -1/2 = -1 + 1/2]
1 + [ -1 + 1/2] [ 1/2^k] =
1 - 1/2^k + (1/2)(1/2^k) =
1 - 1/2^k + 1 / [ 2^k *2] =
1 - 1/2^k + 1/ 2^(k + 1)
+118667 Show that replacing k with k + 1 in 1 - 1/2k gives an expression equivalent to 1 - 1/2k + 1/2k+1
We can also do this with partial fractions and then it will go in the correct direction.
\(1-\frac{1}{2^{k+1}}\\ =1+\;\;\frac{-1}{2*2^k}\\ \mbox{There are some integers A and B such that}\\ =1+\;\;\frac{A}{2^k}+\frac{B}{2^{k+1}}\\ =1+\;\;\frac{2A}{2^{k+1}}+\frac{B}{2^{k+1}}\\ =1+\;\;\frac{2A}{2^{k+1}}+\frac{B}{2^{k+1}}\\ =1+\;\;\frac{2A+B}{2^{k+1}}\\ \qquad SO\\ \qquad 2A+B=-1\\ \qquad \mbox{One solution to this is }\;\;A=-1\;\;and\;\;B=+1\\ =1+\;\;\frac{-1}{2^k}+\frac{1}{2^{k+1}}\\ =1-\;\;\frac{1}{2^k}+\frac{1}{2^{k+1}}\\ \)
[Any A and B such that 2A+B= -1 will also be true ]
