00:04
Hey everybody, i hope you all are having a nice day.
00:07
Today we're solving problem number 10 from chapter 9, section 4 of the textbook.
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In this problem, we are asked to find what p subk plus 1 would be when you have p.
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Square over 2 times k plus 1.
00:34
So for p sub k plus 1, i'll use green color.
00:39
You're simply going to substitute k plus 1 in.
00:44
To each of the k's on the right side of your equation.
00:50
So remember, just put parentheses around whenever you sub in something.
00:53
So you have k plus 1 squared all over two times k plus 1.
01:05
Second parentheses there.
01:08
Just so you guys see where i'm substituting it.
01:10
So substitution is your first step for this problem.
01:15
And you're going to keep adding that one outside.
01:18
Close parentheses and square all that.
01:21
So this simplifies down to k plus 1 squared.
01:30
And now the next few, actually all the steps until your final answer now will be involving simplification.
01:36
We've substituted in and then you've got to simplify.
01:39
So this k plus 1 plus 1 is simply k plus 2 squared.
01:44
And remember you have to rewrite all of this, both the new mirror and denominator in expanded form.
01:51
So k plus 1 square, just to show you all, would look like in its expanded.
01:57
Form again would be k plus 1 times another k plus 1.
02:10
So this would be k times k or k squared...