Math Question and Answers


Question and Answers Forum

All Questions Topic List
None Questions
Previous in All Question Next in All Question
Previous in None Next in None
Question Number 73838 by liki last updated on 16/Nov/19

Commented by liki last updated on 16/Nov/19

$. . I n e e d h e l p p l z; Q n 4 (a) i i, (d) i a n d i i$
Commented by mathmax by abdo last updated on 16/Nov/19
$4 ii) let S=2×3 +3×4 +4×5+...(nterms) ⇒ S=Σ_(k=2) ^(n+1) k(k+1) =Σ_(k=2) ^(n+1) k^2 +Σ_(k=2) ^(n+1) k =Σ_(k=1) ^(n+1) k^2 +Σ_(k=1) ^(n+1) k −2 =(((n+1)(n+2)(2(n+1)+1))/6) +(((n+1)(n+2)/2)−2 =(((n+1)(n+2)(2n+3))/6) +((3(n+1)(n+2))/6) =(((n+1)(n+2))/6){ 2n+3 +3} =(((n+1)(n+2)(2n+6))/6) S=(((n+1)(n+2)(n+3))/3) .$
$4 i i) l e t S = 2 \times 3 + 3 \times 4 + 4 \times 5 + . . . (n t e r m s) \Rightarrow$ $S = \sum_{k = 2}^{n + 1} k (k + 1) = \sum_{k = 2}^{n + 1} k^{2} + \sum_{k = 2}^{n + 1} k$ $= \sum_{k = 1}^{n + 1} k^{2} + \sum_{k = 1}^{n + 1} k - 2 = \frac{(n + 1) (n + 2) (2 (n + 1) + 1)}{6} + \frac{(n + 1) (n + 2}{2} - 2$ $= \frac{(n + 1) (n + 2) (2 n + 3)}{6} + \frac{3 (n + 1) (n + 2)}{6}$ $= \frac{(n + 1) (n + 2)}{6} {2 n + 3 + 3} = \frac{(n + 1) (n + 2) (2 n + 6)}{6}$ $S = \frac{(n + 1) (n + 2) (n + 3)}{3} .$
Commented by mathmax by abdo last updated on 16/Nov/19

$f o r g i v e S = \frac{(n + 1) (n + 2) (n + 3)}{3} - 2 .$
Commented by mathmax by abdo last updated on 16/Nov/19

$4) b) F (x) = \frac{2 x^{2} + 8 x + 7}{(x + 2) (x + 3)} = \frac{2 x^{2} + 8 x + 7}{x^{2} + 5 x + 6}$ $= \frac{2 (x^{2} + 5 x + 6) - 10 x - 12 + 8 x + 7}{x^{2} + 5 x + 6} = 2 - \frac{2 x + 5}{x^{2} + 5 x + 6} l e t$ $f (x) = \frac{2 x + 5}{x^{2} + 5 x + 6} = \frac{2 x + 5}{(x + 2) (x + 3)} = \frac{a}{x + 2} + \frac{b}{x + 3}$ $a = (x + 2) f (x) ∣_{x = - 2} = \frac{1}{1} = 1$ $b = (x + 3) f (x) ∣_{x = - 3} = \frac{- 1}{- 1} = 1 \Rightarrow$ $F (x) = 2 - \frac{1}{x + 2} - \frac{1}{x + 3}$
Commented by mathmax by abdo last updated on 16/Nov/19

$4 d i) w e h a v e {(3 x - \frac{2}{x^{2}})}^{18} = \frac{{(3 x^{3} - 2)}^{18}}{x^{36}}$ $= \frac{1}{x^{36}} \sum_{k = 0}^{18} C_{18}^{k} {(3 x^{3})}^{k} {(- 2)}^{18 - k} = \frac{1}{x^{36}} \sum_{k = 0}^{18} C_{18}^{k} 3^{k} x^{3 k} {(- 2)}^{18 - k}$ $= \sum_{k = 0}^{18} {(- 2)}^{18 - k} 3^{k} C_{18}^{k} x^{3 k - 36} w e g e t t h e i n d e p e n d e n t t e r m i f$ $3 k - 36 = 0 \Rightarrow k = 12 \Rightarrow a_{0} = {(- 2)}^{18 - 12} 3^{12} C_{18}^{12}$ $= {(- 2)}^{6} \times 3^{12} \times C_{18}^{12}$
Commented by liki last updated on 16/Nov/19

$. . . T h a n k s s i r$
Commented by abdomathmax last updated on 16/Nov/19

$y o u a r e w e l c o m e .$
Commented by mathmax by abdo last updated on 16/Nov/19

$M = (\begin{matrix} 1 0 0 \\ x 2 0 \end{matrix})$ $(3 1 1)$ $t h e c a r a c t e r i s t i c p o l y n o m o f M i s$ $d e t (M - α I) = \| \begin{matrix} 1 - α 0 0 \\ x 2 - α 0 \end{matrix} \|$ $∣ 3 1 1 - α ∣$ $= (1 - α) \| \begin{matrix} 2 - α 0 \\ 1 1 - α \end{matrix} \| - x \| \begin{matrix} 0 0 \\ 1 1 - α \end{matrix} \| + 3 \| \begin{matrix} 0 0 \\ 2 - α 0 \end{matrix} \|$ $= (1 - α) (2 - α) (1 - α)$ $= {(α - 1)}^{2} (2 - α) = (α^{2} - 2 α + 1) (2 - α)$ $= 2 α^{2} - 4 α + 2 - α^{3} + 2 α^{2} - α = - α^{3} + 4 α^{2} - 5 α + 2$ $C a y l e r h a m i l t o n t h e o r e m g i v e$ $- M^{3} + 4 M^{2} - 5 M + 2 I = 0 \Rightarrow$ $2 I = M^{3} - 4 M^{2} + 5 M \Rightarrow \frac{1}{2} M (M^{2} - 4 M + 5 I) = I \Rightarrow$ $M^{- 1} = \frac{1}{2} (M^{2} - 4 M + 5 I) r e s t t o f i n i s h t h e c a l c u l u s . . . .$
Commented by mathmax by abdo last updated on 16/Nov/19

$d i i) f (x) = (x - a) (x - b) Q (x) + R (x) d e g R < 2 \Rightarrow$ $R (x) = α x + β$ $f (a) = R (a) = α a + β$ $f (b) = R (b) = α b + β \Rightarrow f (a) - f (b) = (a - b) α \Rightarrow α = \frac{f (a) - f (b)}{a - b}$ $β = f (a) - a α = f (a) - a \frac{f (a) - f (b)}{a - b} = \frac{a f (a) - b f (a) - a f (a) + a f (b)}{a - b}$ $= \frac{a f (b) - b f (a)}{a - b} (w i t h c o n d i t i o n a \neq b) s o t h e r e m i n d e r i s$ $α x + β = \frac{f (a) - f (b)}{a - b} x + \frac{a f (b) - b f (a)}{a - b}$
	Answered by Tanmay chaudhury last updated on 16/Nov/19

	$S = 2 \times 3 + 3 \times 4 \times 4 \times 5 + . . .$ $T_{n} = [2 + (n - 1) \times 1] \times [3 + (n - 1) \times 1]$ $= (n + 1) (n + 2) = n^{2} + 3 n + 2$ $S_{n} = Σ T_{n} = \underset{n = 1}{\sum^{n}} n^{2} + 3 \underset{n = 1}{\sum^{n}} n + 2 \underset{n = 1}{\sum^{n}} 1$ $S_{n} = \frac{n (n + 1) (2 n + 1)}{6} + 3 \times \frac{n (n + 1)}{2} + 2 n$
	Answered by Tanmay chaudhury last updated on 16/Nov/19

	$d (i) l e t (r + 1) t h t e r m i s i n d e p d n d e n t o f x (x^{0})$ $18_{C_{r}} {(3 x)}^{18 - r} \times {(\frac{- 2}{x^{2}})}^{r}$ $18_{C_{r}} \times 3^{18 - r} \times x^{18 - r} \times {(- 1)}^{r} \times 2^{r} \times \frac{1}{x^{2 r}}$ $18_{C_{r}} \times 3^{18 - r} \times x^{18 - 3 r} \times {(- 1)}^{r} \times 2^{r}$ $s o x^{18 - 3 r} = x^{0} \to 3 r = 18 r = 6$ $18_{C_{6}} \times 3^{18 - 6} \times {(- 1)}^{6} \times 2^{6}$ $\frac{18!}{6! 12!} \times 3^{12} \times 2^{6}$
	Commented by liki last updated on 16/Nov/19

	$. . . T h a n k s s o m u c h S i r .$
	Commented by Tanmay chaudhury last updated on 16/Nov/19

	$m o s t w e l c o m e . . .$
	Answered by Tanmay chaudhury last updated on 16/Nov/19

	$l e t f (x) = p (x - a) (x - b) + A (x - a) + B (x - b)$ $f (a) = B (a - b) B = \frac{f (a)}{a - b}$ $f (b) = A (b - a) = - A (a - b) A = \frac{f (b)}{- (a - b)}$ $R e m a i n d e r i s A (x - a) + B (x - b)$ $\frac{f (b)}{- (a - b)} (x - a) + \frac{f (a)}{a - b} (x - b)$ $= \frac{f (b) x}{- (a - b)} + \frac{a f (b)}{(a - b)} + \frac{x f (a)}{a - b} - \frac{b f (a)}{a - b}$ $= (\frac{f (a)}{a - b} - \frac{f (b)}{a - b}) x + \frac{a f (b) - b f (a)}{a - b}$ $= (\frac{f (a) - f (b)}{a - b}) x + \frac{a f (b) - b f (a)}{a - b} A n s w e r$
	Commented by liki last updated on 16/Nov/19

	$G o d b l e s s y o u s i r .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum