let give P_n (x)=(x+1)^(2n) +(x+2)^n −1 and Q(x)= x^2 +3x +2 find R(x) /P


Question and Answers Forum

All Questions Topic List
Algebra Questions
Previous in All Question Next in All Question
Previous in Algebra Next in Algebra
Question Number 28312 by abdo imad last updated on 23/Jan/18

$l e t g i v e P_{n} (x) = {(x + 1)}^{2 n} + {(x + 2)}^{n} - 1 a n d$ $Q (x) = x^{2} + 3 x + 2 f i n d R (x) / P_{n} (x) = R (x) Q (x) .$
	Answered by sma3l2996 last updated on 24/Jan/18

	$P (x) = R (x) Q (x) s o R (x) = \frac{P (x)}{Q (x)}$ $R (x) = \frac{{(x + 1)}^{2 n}}{x^{2} + 3 x + 2} + \frac{{(x + 2)}^{n}}{x^{2} + 3 x + 2} - \frac{1}{x^{2} + 3 x + 2}$ $w e h a v e x^{2} + 3 x + 2 = (x + 1) (x + 2)$ $s o R (x) = \frac{{(x + 1)}^{2 n - 1}}{x + 2} + \frac{{(x + 2)}^{n - 1}}{x + 1} - \frac{1}{(x + 1) (x + 2)}$ ${l e t}^{'} s p u t t = x + 1 s o x + 2 = t + 1$ $s o t^{2 n - 1} = (t^{2 n - 2} - t^{2 n - 3} + t^{2 n - 4} - . . . + 1) (t + 1) - 1$ $s o {(x + 1)}^{2 n - 1} = (\underset{k = 2}{\sum^{2 n}} {(- 1)}^{k} {(x + 1)}^{2 n - k}) (x + 2) - 1$ $\frac{{(x + 1)}^{2 n - 1}}{x + 2} = \underset{k = 2}{\sum^{2 n}} {(- 1)}^{k} {(x + 1)}^{2 n - k} - \frac{1}{x + 2}$ $s a m e t h i n g f o r \frac{{(x + 2)}^{n - 1}}{x + 1} = \underset{k = 2}{\sum^{n}} {(x + 2)}^{n - k} + \frac{1}{x + 1}$ $s o R (x) = \underset{k = 2}{\sum^{2 n}} {(- 1)}^{k} {(x + 1)}^{2 n - k} + \underset{k = 2}{\sum^{n}} {(x + 2)}^{n - k} + \frac{1}{x + 1} - \frac{1}{x + 2} - \frac{1}{(x + 1) (x + 2)}$ $\frac{1}{(x + 2) (x + 1)} = \frac{1}{x + 1} - \frac{1}{x + 2}$ $s o$ $R (x) = \underset{k = 2}{\sum^{n}} {(x + 2)}^{n - k} + \underset{k = 2}{\sum^{2 n}} {(- 1)}^{k} {(x + 1)}^{2 n - k}$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum