induction (1 Viewer)

rainnwind · Feb 9, 2009

Sorry I am having trouble in doing the last part of the question [prove true n=k+1] can anyone help please?

1) see attached img
2) 1+(1+x) + (1+x)^2 + ... + (1+x)^n = [(1+x)^(n+1) -1]/x
3) 1/(x^n)(x-1) = 1/(x-1) - 1/(x^2)(x-1) - ... - 1/(x^n)

kurt.physics · Feb 9, 2009

rainnwind said:
1) $\sum_{r=1}^{n}r(2^r) = (n-1) 2^{n+1} + 2$

$\sum_{r=1}^{n}r(2^r) = (n-1) 2^{n+1} + 2$

test for $n = 1,$

$LHS = 1(2^1)$

$= 2$

$RHS = (1-1)2^{1 + 1} + 2$

$= 0 \times 2^2 + 2$

$= 2$

$\therefore LHS = RHS$

assume true for $n = k$

$2 + 8 + . . . + k(2^k) = (k-1)2^{k+1} + 2$

prove for $n = k + 1$

$2 + 8 + . . . + k(2^k) + (k + 1)2^{k + 1} = (k + 1 - 1)2^{k + 1 + 1} + 2$

$2 + 8 + . . . + k(2^k) + (k + 1)2^{k + 1} = k \times 2^{k + 2} + 2$

$LHS = (k-1)2^{k+1} + 2 + (k + 1)2^{k + 1}$

rearranging,

$= 2^{k+1}(k-1) + 2^{k+1}(k + 1) + 2$

factoring $2^{k+1}$

$= 2^{k+1}( (k-1) + (k + 1) ) + 2$

$= 2^{k+1}( (k - 1 + k + 1) ) + 2$

$= 2^{k+1}(2k) + 2$

$= 2^{k+1} \times 2^1 \times k + 2$

$= 2^{k+ 1 + 1} \times k + 2$

$= k \times 2^{k+ 2} + 2$

$= RHS$

$\therefore$ if the formula is true for $n = k$ , then it is true for $n = k + 1$

We know that the formula is true for $n = 1$ , so it must be true for $n = 2$ . If we know it is true for $n = 2$ , then it must be true for $n = 3$ , and so on.

$\therefore$ it is true for all $n > 1$

kurt.physics · Feb 9, 2009

rainnwind said:
2) $1 + (1+x) + (1+x)^2 + . . . + (1+x)^n = \frac{(1 + x)^{n+1} - 1}{x}$

$1 + (1+x) + (1+x)^2 + . . . + (1+x)^n = \frac{(1 + x)^{n+1} - 1}{x}$

test for $n = 1,$

$LHS = 1 + (1 + x)^1$

$= 1 + 1 + x$

$= 2 + x$

$RHS = \frac{(1 + x)^{2} - 1}{x}$

$= \frac{(1 + x + 1)(1 + x - 1)}{x}$

$= \frac{(x + 2)x}{x}$

$= 2 + x$

$\therefore LHS = RHS$

assume true for $n = k$

$1 + (1+x) + (1+x)^2 + . . . + (1+x)^k = \frac{(1 + x)^{k+1} - 1}{x}$

prove for $n = k + 1$

$1 + (1+x) + (1+x)^2 + . . . + (1+x)^k + (1+x)^{k+1} = \frac{(1 + x)^{k+1 + 1} - 1}{x}$

$1 + (1+x) + (1+x)^2 + . . . + (1+x)^k + (1+x)^{k+1} = \frac{(1 + x)^{k+2} - 1}{x}$

$LHS = 1 + (1+x) + (1+x)^2 + . . . + (1+x)^k + (1+x)^{k+1}$ [\

$= \frac{(1 + x)^{k+1} - 1}{x} + (1+x)^{k+1}$

$= \frac{(1 + x)^{k+1} - 1}{x} +\frac{x( (1+x)^{k+1} )}{x}$

$= \frac{ (1+x)^{k+1} + x(1+x)^{k+1} - 1}{x}$

factoring $(1+x)^{k+1}$ ,

$= \frac{ (1+x)^{k+1}(1 + x) - 1}{x}$

$= \frac{ (1+x)^{k+1}(x + 1)^1 - 1}{x}$

$= \frac{ (1+x)^{k+1+1} - 1}{x}$

$= \frac{ (1+x)^{k+2} - 1}{x}$

$= RHS$

$\therefore$ if the formula is true for $n = k$ , then it is true for $n = k + 1$

We know that the formula is true for $n = 1$ , so it must be true for $n = 2$ . If we know it is true for $n = 2$ , then it must be true for $n = 3$ , and so on.

$\therefore$ it is true for all $n > 1$

Trebla · Feb 9, 2009

The 2nd one is just a geometric series lol

Check your 3rd question...it doesn't look right to me...

kurt.physics · Feb 9, 2009

Trebla said:
Check your 3rd question...it doesn't look right to me...

Yeh, i also though that question 3 looked wrong to me.

rainnwind · Feb 9, 2009

thank you so much... [i know how to do it, but silly mistake... i knew i was stuck because i made some errors when proving but i wasn't sure where were they when it's so obvious...=.=]

the last question is printed as it is on the sheet, i thought it was me who kept making mistakes in the calculation... thanks.

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