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AlgebraQuestion and Answers: Page 352

Question Number 23566 Answers: 1 Comments: 2

((a/b))^(log c) .((b/c))^(log a) .((c/a))^(log b) =1. make ((logc)/c) the subject of formula

$$\left(\frac{{a}}{{b}}\right)^{\mathrm{log}\:{c}} .\left(\frac{{b}}{{c}}\right)^{\mathrm{log}\:{a}} .\left(\frac{{c}}{{a}}\right)^{\mathrm{log}\:{b}} =\mathrm{1}. \\ $$$$ \\ $$$${make}\:\frac{{logc}}{{c}}\:{the}\:{subject}\:{of}\:{formula} \\ $$

Question Number 23537 Answers: 1 Comments: 8

Question Number 23479 Answers: 0 Comments: 0

Common solution. (d/dy)(u_x +u)+2x^2 y(u_x +u)=0.

$$\boldsymbol{\mathrm{Common}}\:\:\boldsymbol{\mathrm{solution}}. \\ $$$$\frac{\boldsymbol{\mathfrak{d}}}{\boldsymbol{\mathfrak{d}\mathrm{y}}}\left(\boldsymbol{\mathrm{u}}_{\boldsymbol{\mathrm{x}}} +\boldsymbol{\mathrm{u}}\right)+\mathrm{2}\boldsymbol{\mathrm{x}}^{\mathrm{2}} \boldsymbol{\mathrm{y}}\left(\boldsymbol{\mathrm{u}}_{\boldsymbol{\mathrm{x}}} +\boldsymbol{\mathrm{u}}\right)=\mathrm{0}. \\ $$

Question Number 23471 Answers: 1 Comments: 0

Prove that ΣΣ_(0≤i<j≤n) ((1/(^n C_i )) + (1/(^n C_j ))) = Σ_(r=0) ^(n−1) ((n − r)/(^n C_r )) + Σ_(r=1) ^n (r/(^n C_r ))

$${Prove}\:{that} \\ $$$$\underset{\mathrm{0}\leqslant{i}<{j}\leqslant{n}} {\Sigma\Sigma}\left(\frac{\mathrm{1}}{\:^{{n}} {C}_{{i}} }\:+\:\frac{\mathrm{1}}{\:^{{n}} {C}_{{j}} }\right)\:=\:\underset{{r}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\frac{{n}\:−\:{r}}{\:^{{n}} {C}_{{r}} }\:+\:\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{{r}}{\:^{{n}} {C}_{{r}} } \\ $$

Question Number 23357 Answers: 0 Comments: 0

Prove that (1/(m!))C_0 +(n/((m+1)!))C_1 +((n(n−1))/((m+2)!))C_2 +...+((n(n−1)...2.1)/((m+n)!))C_n = (((m+n+1)(m+n+2)...(m+2n))/((m+n)!)).

$${Prove}\:{that}\:\frac{\mathrm{1}}{{m}!}{C}_{\mathrm{0}} +\frac{{n}}{\left({m}+\mathrm{1}\right)!}{C}_{\mathrm{1}} +\frac{{n}\left({n}−\mathrm{1}\right)}{\left({m}+\mathrm{2}\right)!}{C}_{\mathrm{2}} \\ $$$$+...+\frac{{n}\left({n}−\mathrm{1}\right)...\mathrm{2}.\mathrm{1}}{\left({m}+{n}\right)!}{C}_{{n}} = \\ $$$$\frac{\left({m}+{n}+\mathrm{1}\right)\left({m}+{n}+\mathrm{2}\right)...\left({m}+\mathrm{2}{n}\right)}{\left({m}+{n}\right)!}. \\ $$

Question Number 23334 Answers: 1 Comments: 3

Question Number 23323 Answers: 0 Comments: 1

Question Number 23250 Answers: 0 Comments: 0

If (1 + x)^n = C_0 + C_1 x + C_2 x^2 + C_3 x^3 + ... + C_n x^n , then prove that ΣΣ_(0≤i<j≤n) ((i/(^n C_i )) + (j/(^n C_j ))) = (n^2 /2)(Σ_(r=0) ^n (1/(^n C_r ))).

$${If}\:\left(\mathrm{1}\:+\:{x}\right)^{{n}} \:=\:{C}_{\mathrm{0}} \:+\:{C}_{\mathrm{1}} {x}\:+\:{C}_{\mathrm{2}} {x}^{\mathrm{2}} \:+\:{C}_{\mathrm{3}} {x}^{\mathrm{3}} \\ $$$$+\:...\:+\:{C}_{{n}} {x}^{{n}} ,\:{then}\:{prove}\:{that} \\ $$$$\underset{\mathrm{0}\leqslant{i}<{j}\leqslant{n}} {\Sigma\Sigma}\left(\frac{{i}}{\:^{{n}} {C}_{{i}} }\:+\:\frac{{j}}{\:^{{n}} {C}_{{j}} }\right)\:=\:\frac{{n}^{\mathrm{2}} }{\mathrm{2}}\left(\underset{{r}=\mathrm{0}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{\:^{{n}} {C}_{{r}} }\right). \\ $$

Question Number 23208 Answers: 1 Comments: 0

Is it possible to find how many real roots exist in the equation x^4 + ∣x∣ = 3 without find all the value of x?

$$\mathrm{Is}\:\mathrm{it}\:\mathrm{possible}\:\mathrm{to}\:\mathrm{find}\:\mathrm{how}\:\mathrm{many}\:\mathrm{real}\:\mathrm{roots}\: \\ $$$$\mathrm{exist}\:\mathrm{in}\:\mathrm{the}\:\mathrm{equation} \\ $$$${x}^{\mathrm{4}} \:+\:\mid{x}\mid\:=\:\mathrm{3} \\ $$$$\mathrm{without}\:\mathrm{find}\:\mathrm{all}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\:{x}? \\ $$

Question Number 23096 Answers: 0 Comments: 1

Solve: 3^(x ) = ((27)/x) + 18

$$\mathrm{Solve}:\:\:\mathrm{3}^{\mathrm{x}\:} =\:\frac{\mathrm{27}}{\mathrm{x}}\:+\:\mathrm{18} \\ $$

Question Number 23000 Answers: 0 Comments: 0

Let n be a positive integer and p_1 , p_2 , ..., p_n be n prime numbers all larger than 5 such that 6 divides p_1 ^2 + p_2 ^2 + ... + p_n ^2 . Prove that 6 divides n.

$$\mathrm{Let}\:{n}\:\mathrm{be}\:\mathrm{a}\:\mathrm{positive}\:\mathrm{integer}\:\mathrm{and}\:{p}_{\mathrm{1}} ,\:{p}_{\mathrm{2}} , \\ $$$$...,\:{p}_{{n}} \:\mathrm{be}\:{n}\:\mathrm{prime}\:\mathrm{numbers}\:\mathrm{all}\:\mathrm{larger} \\ $$$$\mathrm{than}\:\mathrm{5}\:\mathrm{such}\:\mathrm{that}\:\mathrm{6}\:\mathrm{divides}\:{p}_{\mathrm{1}} ^{\mathrm{2}} \:+\:{p}_{\mathrm{2}} ^{\mathrm{2}} \:+\:...\:+ \\ $$$${p}_{{n}} ^{\mathrm{2}} .\:\mathrm{Prove}\:\mathrm{that}\:\mathrm{6}\:\mathrm{divides}\:{n}. \\ $$

Question Number 22945 Answers: 0 Comments: 0

Question Number 22856 Answers: 0 Comments: 1

solve: sin(x)=2, x∈C

$$\mathrm{solve}:\:\mathrm{sin}\left({x}\right)=\mathrm{2},\:\:\:{x}\in\mathbb{C} \\ $$

Question Number 22768 Answers: 1 Comments: 2

Question Number 22739 Answers: 0 Comments: 0

If (1 + x)^n = C_0 + C_1 x + C_2 x^2 + C_3 x^3 + ... + C_n x^n , Prove that ΣΣ_(0≤i<j≤n) (i + j)C_i C_j = n(2^(2n−1) − (1/2)^(2n) C_n )

Question Number 22701 Answers: 2 Comments: 0

Question Number 22618 Answers: 1 Comments: 0

If (1 + x)^n = C_0 + C_1 x + C_2 x^2 + C_3 x^3 + ... + C_n x^n , prove that (2^2 /(1.2))C_0 + (2^3 /(2.3))C_1 + (2^4 /(3.4))C_2 + ... + (2^(n+2) /((n + 1)(n + 2)))C_n = ((3^(n+2) − 2n − 5)/((n + 1)(n + 2)))

$${If}\:\left(\mathrm{1}\:+\:{x}\right)^{{n}} \:=\:{C}_{\mathrm{0}} \:+\:{C}_{\mathrm{1}} {x}\:+\:{C}_{\mathrm{2}} {x}^{\mathrm{2}} \:+\:{C}_{\mathrm{3}} {x}^{\mathrm{3}} \\ $$$$+\:...\:+\:{C}_{{n}} {x}^{{n}} ,\:{prove}\:{that} \\ $$$$\frac{\mathrm{2}^{\mathrm{2}} }{\mathrm{1}.\mathrm{2}}{C}_{\mathrm{0}} \:+\:\frac{\mathrm{2}^{\mathrm{3}} }{\mathrm{2}.\mathrm{3}}{C}_{\mathrm{1}} \:+\:\frac{\mathrm{2}^{\mathrm{4}} }{\mathrm{3}.\mathrm{4}}{C}_{\mathrm{2}} \:+\:...\:+ \\ $$$$\frac{\mathrm{2}^{{n}+\mathrm{2}} }{\left({n}\:+\:\mathrm{1}\right)\left({n}\:+\:\mathrm{2}\right)}{C}_{{n}} \:=\:\frac{\mathrm{3}^{{n}+\mathrm{2}} \:−\:\mathrm{2}{n}\:−\:\mathrm{5}}{\left({n}\:+\:\mathrm{1}\right)\left({n}\:+\:\mathrm{2}\right)} \\ $$

Question Number 22640 Answers: 0 Comments: 0

With usual notation, show that (C_0 /x) − (C_1 /(x+1)) + (C_2 /(x+2)) − .... + (−1)^n (C_n /(x+n))= ((n!)/(x(x + 1)(x + 2)....(x + n)))

$${With}\:{usual}\:{notation},\:{show}\:{that} \\ $$$$\frac{{C}_{\mathrm{0}} }{{x}}\:−\:\frac{{C}_{\mathrm{1}} }{{x}+\mathrm{1}}\:+\:\frac{{C}_{\mathrm{2}} }{{x}+\mathrm{2}}\:−\:....\:+\:\left(−\mathrm{1}\right)^{{n}} \frac{{C}_{{n}} }{{x}+{n}}= \\ $$$$\frac{{n}!}{{x}\left({x}\:+\:\mathrm{1}\right)\left({x}\:+\:\mathrm{2}\right)....\left({x}\:+\:{n}\right)} \\ $$

Question Number 22612 Answers: 2 Comments: 0

In the binomial expasion of (a − b)^5 , the sum of 2^(nd) and 3^(rd) term is zero, then (a/b) is

$${In}\:{the}\:{binomial}\:{expasion}\:{of}\:\left({a}\:−\:{b}\right)^{\mathrm{5}} , \\ $$$${the}\:{sum}\:{of}\:\mathrm{2}^{{nd}} \:{and}\:\mathrm{3}^{{rd}} \:{term}\:{is}\:{zero}, \\ $$$${then}\:\frac{{a}}{{b}}\:{is} \\ $$

Question Number 22547 Answers: 1 Comments: 0

If α = (5/(2!3)) + ((5.7)/(3!3^2 )) + ((5.7.9)/(4!3^3 )) ,... then find the value of α^2 + 4α.

$$\mathrm{If}\:\alpha\:=\:\frac{\mathrm{5}}{\mathrm{2}!\mathrm{3}}\:+\:\frac{\mathrm{5}.\mathrm{7}}{\mathrm{3}!\mathrm{3}^{\mathrm{2}} }\:+\:\frac{\mathrm{5}.\mathrm{7}.\mathrm{9}}{\mathrm{4}!\mathrm{3}^{\mathrm{3}} }\:,...\:\mathrm{then}\:\mathrm{find} \\ $$$$\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\alpha^{\mathrm{2}} \:+\:\mathrm{4}\alpha. \\ $$

Question Number 22517 Answers: 0 Comments: 0

Find the coefficient of x in the expansion of [(√(1 + x^2 )) − x]^(−1) in ascending power of x when ∣x∣ < 1.

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{coefficient}\:\mathrm{of}\:{x}\:\mathrm{in}\:\mathrm{the}\:\mathrm{expansion} \\ $$$$\mathrm{of}\:\left[\sqrt{\mathrm{1}\:+\:{x}^{\mathrm{2}} }\:−\:{x}\right]^{−\mathrm{1}} \:\mathrm{in}\:\mathrm{ascending}\:\mathrm{power} \\ $$$$\mathrm{of}\:{x}\:\mathrm{when}\:\mid{x}\mid\:<\:\mathrm{1}. \\ $$

Question Number 22503 Answers: 1 Comments: 4

If (a + bx)^(−2) = (1/4) − 3x + ..., then (a, b) =

$$\mathrm{If}\:\left({a}\:+\:{bx}\right)^{−\mathrm{2}} \:=\:\frac{\mathrm{1}}{\mathrm{4}}\:−\:\mathrm{3}{x}\:+\:...,\:\mathrm{then}\:\left({a},\:{b}\right)\:= \\ $$

Question Number 22491 Answers: 1 Comments: 0

The coefficient of x^r in the expansion of (1 − 2x)^(−1/2) is (1) (((2r)!)/((r!)^2 )) (2) (((2r)!)/(2^r (r!)^2 )) (3) (((2r)!)/((r!)^2 2^(2r) )) (4) (((2r)!)/(2^r (r + 1)!(r − 1)!))

$$\mathrm{The}\:\mathrm{coefficient}\:\mathrm{of}\:{x}^{{r}} \:\mathrm{in}\:\mathrm{the}\:\mathrm{expansion}\:\mathrm{of} \\ $$$$\left(\mathrm{1}\:−\:\mathrm{2}{x}\right)^{−\mathrm{1}/\mathrm{2}} \:\mathrm{is} \\ $$$$\left(\mathrm{1}\right)\:\frac{\left(\mathrm{2}{r}\right)!}{\left({r}!\right)^{\mathrm{2}} } \\ $$$$\left(\mathrm{2}\right)\:\frac{\left(\mathrm{2}{r}\right)!}{\mathrm{2}^{{r}} \:\left({r}!\right)^{\mathrm{2}} } \\ $$$$\left(\mathrm{3}\right)\:\frac{\left(\mathrm{2}{r}\right)!}{\left({r}!\right)^{\mathrm{2}} \:\mathrm{2}^{\mathrm{2}{r}} } \\ $$$$\left(\mathrm{4}\right)\:\frac{\left(\mathrm{2}{r}\right)!}{\mathrm{2}^{{r}} \:\left({r}\:+\:\mathrm{1}\right)!\left({r}\:−\:\mathrm{1}\right)!} \\ $$

Question Number 22474 Answers: 0 Comments: 0

Let R = (5(√5) + 11)^(2n+1) and f = R − [R], then prove that Rf = 4^(2n+1) .

$$\mathrm{Let}\:{R}\:=\:\left(\mathrm{5}\sqrt{\mathrm{5}}\:+\:\mathrm{11}\right)^{\mathrm{2}{n}+\mathrm{1}} \:\mathrm{and}\:{f}\:=\:{R}\:−\:\left[{R}\right], \\ $$$$\mathrm{then}\:\mathrm{prove}\:\mathrm{that}\:{Rf}\:=\:\mathrm{4}^{\mathrm{2}{n}+\mathrm{1}} . \\ $$

Question Number 22472 Answers: 0 Comments: 0

If x^x ∙y^y ∙z^z = x^y ∙y^z ∙z^x = x^z ∙y^x ∙z^y such that x, y and z are positive integers greater than 1, then which of the following cannot be true for any of the possible value of x, y and z? (1) xyz = 27 (2) xyz = 1728 (3) x + y + z = 32 (4) x + y + z = 12

$$\mathrm{If}\:{x}^{{x}} \centerdot{y}^{{y}} \centerdot{z}^{{z}} \:=\:{x}^{{y}} \centerdot{y}^{{z}} \centerdot{z}^{{x}} \:=\:{x}^{{z}} \centerdot{y}^{{x}} \centerdot{z}^{{y}} \:\mathrm{such} \\ $$$$\mathrm{that}\:{x},\:{y}\:\mathrm{and}\:{z}\:\mathrm{are}\:\mathrm{positive}\:\mathrm{integers} \\ $$$$\mathrm{greater}\:\mathrm{than}\:\mathrm{1},\:\mathrm{then}\:\mathrm{which}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{following}\:\mathrm{cannot}\:\mathrm{be}\:\mathrm{true}\:\mathrm{for}\:\mathrm{any}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{possible}\:\mathrm{value}\:\mathrm{of}\:{x},\:{y}\:\mathrm{and}\:{z}? \\ $$$$\left(\mathrm{1}\right)\:{xyz}\:=\:\mathrm{27} \\ $$$$\left(\mathrm{2}\right)\:{xyz}\:=\:\mathrm{1728} \\ $$$$\left(\mathrm{3}\right)\:{x}\:+\:{y}\:+\:{z}\:=\:\mathrm{32} \\ $$$$\left(\mathrm{4}\right)\:{x}\:+\:{y}\:+\:{z}\:=\:\mathrm{12} \\ $$

Question Number 22468 Answers: 0 Comments: 0

If a_r is the coefficient of x^r in the expansion (1 + x + x^2 )^n , then a_1 − 2a_2 + 3a_3 − ....... 2na_(2n) =

$$\mathrm{If}\:{a}_{{r}} \:\mathrm{is}\:\mathrm{the}\:\mathrm{coefficient}\:\mathrm{of}\:{x}^{{r}} \:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{expansion}\:\left(\mathrm{1}\:+\:{x}\:+\:{x}^{\mathrm{2}} \right)^{{n}} ,\:\mathrm{then} \\ $$$${a}_{\mathrm{1}} \:−\:\mathrm{2}{a}_{\mathrm{2}} \:+\:\mathrm{3}{a}_{\mathrm{3}} \:−\:.......\:\mathrm{2}{na}_{\mathrm{2}{n}} \:= \\ $$

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