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AlgebraQuestion and Answers: Page 352

Question Number 24542 Answers: 0 Comments: 2

Prove that coefficient of x^n in ((a+bx+cx^2 )/e^x ) is (((−1)^n )/(n!))[cn^2 −(b+c)n+a]

$${Prove}\:{that}\:{coefficient}\:{of}\:{x}^{{n}} \:{in} \\ $$$$\frac{{a}+{bx}+{cx}^{\mathrm{2}} }{{e}^{{x}} }\:{is}\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!}\left[{cn}^{\mathrm{2}} −\left({b}+{c}\right){n}+{a}\right] \\ $$

Question Number 24540 Answers: 2 Comments: 1

Prove that (i) Σ_(n=0) ^∞ (n^2 /(n!))=2e. (ii) Σ_(n=0) ^∞ (n^3 /(n!))=5e. (iii) Σ_(n=0) ^∞ (n^4 /(n!))=15e.

$${Prove}\:{that} \\ $$$$\left({i}\right)\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{n}^{\mathrm{2}} }{{n}!}=\mathrm{2}{e}. \\ $$$$\left({ii}\right)\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{n}^{\mathrm{3}} }{{n}!}=\mathrm{5}{e}. \\ $$$$\left({iii}\right)\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{n}^{\mathrm{4}} }{{n}!}=\mathrm{15}{e}. \\ $$

Question Number 24469 Answers: 0 Comments: 2

Let 2x + 3y + 4z = 9, x, y, z > 0 then the maximum value of (1 + x)^2 (2 + y)^3 (4 + z)^4 is

$$\mathrm{Let}\:\mathrm{2}{x}\:+\:\mathrm{3}{y}\:+\:\mathrm{4}{z}\:=\:\mathrm{9},\:{x},\:{y},\:{z}\:>\:\mathrm{0}\:\mathrm{then} \\ $$$$\mathrm{the}\:\mathrm{maximum}\:\mathrm{value}\:\mathrm{of}\:\left(\mathrm{1}\:+\:{x}\right)^{\mathrm{2}} \:\left(\mathrm{2}\:+\:{y}\right)^{\mathrm{3}} \\ $$$$\left(\mathrm{4}\:+\:{z}\right)^{\mathrm{4}} \:\mathrm{is} \\ $$

Question Number 24443 Answers: 2 Comments: 0

Solve for x: (10^(−4) x)^x =4×10^(−8)

$${Solve}\:{for}\:{x}: \\ $$$$\left(\mathrm{10}^{−\mathrm{4}} {x}\right)^{{x}} =\mathrm{4}×\mathrm{10}^{−\mathrm{8}} \\ $$

Question Number 24438 Answers: 0 Comments: 2

Find the sum to infinite terms of the series (x/(1−x^2 ))+(x^2 /(1−x^4 ))+(x^4 /(1−x^8 ))+......

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{to}\:\mathrm{infinite}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{series}\:\frac{{x}}{\mathrm{1}−{x}^{\mathrm{2}} }+\frac{{x}^{\mathrm{2}} }{\mathrm{1}−{x}^{\mathrm{4}} }+\frac{{x}^{\mathrm{4}} }{\mathrm{1}−{x}^{\mathrm{8}} }+...... \\ $$

Question Number 24360 Answers: 0 Comments: 1

Question Number 24359 Answers: 1 Comments: 1

Question Number 24286 Answers: 1 Comments: 0

If ∣x∣ < 1 then (x + 1)(x^2 + 1)(x^4 + 1)(x^8 + 1)(x^(16) + 1)..... is equal to

$$\mathrm{If}\:\mid{x}\mid\:<\:\mathrm{1}\:\mathrm{then} \\ $$$$\left({x}\:+\:\mathrm{1}\right)\left({x}^{\mathrm{2}} \:+\:\mathrm{1}\right)\left({x}^{\mathrm{4}} \:+\:\mathrm{1}\right)\left({x}^{\mathrm{8}} \:+\:\mathrm{1}\right)\left({x}^{\mathrm{16}} \:+\:\mathrm{1}\right)..... \\ $$$$\mathrm{is}\:\mathrm{equal}\:\mathrm{to} \\ $$

Question Number 24142 Answers: 0 Comments: 2

Prove that Σ_(r=1) ^(2n−1) (−1)^(r−1) (∫_0 ^1 x^r (1−x)^(2n−r) dx) =∫_0 ^1 [(1−x)^(2n) +x^(2n) −(1−x)^(2n+1) −x^(2n+1) ]dx

$${Prove}\:{that} \\ $$$$\underset{{r}=\mathrm{1}} {\overset{\mathrm{2}{n}−\mathrm{1}} {\sum}}\left(−\mathrm{1}\right)^{{r}−\mathrm{1}} \left(\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}{x}^{{r}} \left(\mathrm{1}−{x}\right)^{\mathrm{2}{n}−{r}} {dx}\right) \\ $$$$=\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\left[\left(\mathrm{1}−{x}\right)^{\mathrm{2}{n}} +{x}^{\mathrm{2}{n}} −\left(\mathrm{1}−{x}\right)^{\mathrm{2}{n}+\mathrm{1}} −{x}^{\mathrm{2}{n}+\mathrm{1}} \right]{dx} \\ $$

Question Number 24069 Answers: 1 Comments: 0

Simplify (((log_2 (√5) . log_(25) 20) + log_4 (√(50)) )/(log_4 70 − log_(15) 49))

$$\mathrm{Simplify} \\ $$$$\frac{\left(\mathrm{log}_{\mathrm{2}} \:\sqrt{\mathrm{5}}\:.\:\mathrm{log}_{\mathrm{25}} \:\mathrm{20}\right)\:+\:\mathrm{log}_{\mathrm{4}} \:\sqrt{\mathrm{50}}\:\:}{\mathrm{log}_{\mathrm{4}} \:\mathrm{70}\:−\:\mathrm{log}_{\mathrm{15}} \:\mathrm{49}} \\ $$

Question Number 24010 Answers: 0 Comments: 0

Prove that Σ_(r=1) ^(2n−1) (−1)^(r−1) ∙(r/(^(2n) C_r )) = (n/(n + 1)) .

$$\mathrm{Prove}\:\mathrm{that}\:\underset{{r}=\mathrm{1}} {\overset{\mathrm{2}{n}−\mathrm{1}} {\sum}}\left(−\mathrm{1}\right)^{{r}−\mathrm{1}} \centerdot\frac{{r}}{\:^{\mathrm{2}{n}} {C}_{{r}} }\:=\:\frac{{n}}{{n}\:+\:\mathrm{1}}\:. \\ $$

Question Number 23968 Answers: 0 Comments: 3

Prove that n and 2n−1 are coprime.

$${Prove}\:{that}\:{n}\:{and}\:\mathrm{2}{n}−\mathrm{1}\:{are}\:{coprime}. \\ $$

Question Number 23953 Answers: 0 Comments: 4

find the nth term of the sequence 4, 9, 16, 45, 76 please help

$${find}\:{the}\:{nth}\:{term}\:{of}\:{the}\:{sequence} \\ $$$$\mathrm{4},\:\mathrm{9},\:\mathrm{16},\:\mathrm{45},\:\mathrm{76} \\ $$$$ \\ $$$${please}\:{help} \\ $$

Question Number 23928 Answers: 0 Comments: 4

Find^n C_1 − (1/2)^n C_2 + (1/3)^n C_3 − .... + (−1)^(n−1) (1/n) ∙^n C_n .

$$\mathrm{Find}\:^{{n}} {C}_{\mathrm{1}} \:−\:\frac{\mathrm{1}}{\mathrm{2}}\:^{{n}} {C}_{\mathrm{2}} \:+\:\frac{\mathrm{1}}{\mathrm{3}}\:^{{n}} {C}_{\mathrm{3}} \:−\:....\:+ \\ $$$$\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \frac{\mathrm{1}}{{n}}\:\centerdot\:^{{n}} {C}_{{n}} . \\ $$

Question Number 23884 Answers: 1 Comments: 0

If C_r stands for^n C_r = ((n!)/(r! n − r!)) and Σ_(r=1) ^n r.C_r ^2 = λ for n ≥ 2, then λ is divisible by (1) 3 (n − 1) (2) n + 1 (3) n (2n − 1) (4) n^2 + 1

$$\mathrm{If}\:{C}_{{r}} \:\mathrm{stands}\:\mathrm{for}\:^{{n}} {C}_{{r}} \:=\:\frac{{n}!}{{r}!\:{n}\:−\:{r}!}\:\mathrm{and} \\ $$$$\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}{r}.{C}_{{r}} ^{\mathrm{2}} \:=\:\lambda\:\mathrm{for}\:{n}\:\geqslant\:\mathrm{2},\:\mathrm{then}\:\lambda\:\mathrm{is}\:\mathrm{divisible} \\ $$$$\mathrm{by} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{3}\:\left({n}\:−\:\mathrm{1}\right) \\ $$$$\left(\mathrm{2}\right)\:{n}\:+\:\mathrm{1} \\ $$$$\left(\mathrm{3}\right)\:{n}\:\left(\mathrm{2}{n}\:−\:\mathrm{1}\right) \\ $$$$\left(\mathrm{4}\right)\:{n}^{\mathrm{2}} \:+\:\mathrm{1} \\ $$

Question Number 23814 Answers: 1 Comments: 1

Question Number 23700 Answers: 0 Comments: 3

If (1 − x^3 )^n = Σ_(r=0) ^n a_r x^r (1 − x)^(3n−2r) , then the value of a_r , where n ∈ N is (1)^n C_r ∙3^r (2)^n C_(3r) (3)^n C_(r−1) 2^(r−1) (4)^n C_r 2^r

$$\mathrm{If}\:\left(\mathrm{1}\:−\:{x}^{\mathrm{3}} \right)^{{n}} \:=\:\underset{{r}=\mathrm{0}} {\overset{{n}} {\sum}}{a}_{{r}} {x}^{{r}} \left(\mathrm{1}\:−\:{x}\right)^{\mathrm{3}{n}−\mathrm{2}{r}} ,\:\mathrm{then} \\ $$$$\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{a}_{{r}} ,\:\mathrm{where}\:{n}\:\in\:{N}\:\mathrm{is} \\ $$$$\left(\mathrm{1}\right)\:^{{n}} {C}_{{r}} \centerdot\mathrm{3}^{{r}} \\ $$$$\left(\mathrm{2}\right)\:^{{n}} {C}_{\mathrm{3}{r}} \\ $$$$\left(\mathrm{3}\right)\:^{{n}} {C}_{{r}−\mathrm{1}} \mathrm{2}^{{r}−\mathrm{1}} \\ $$$$\left(\mathrm{4}\right)\:^{{n}} {C}_{{r}} \mathrm{2}^{{r}} \\ $$

Question Number 23687 Answers: 1 Comments: 0

Solve for x, y and z x^2 − yz = 1 .......... (i) y^2 − xz = 4 .......... (i) z^2 − xy = 9 .......... (i)

$$\mathrm{Solve}\:\mathrm{for}\:\:\mathrm{x},\:\mathrm{y}\:\mathrm{and}\:\mathrm{z} \\ $$$$\mathrm{x}^{\mathrm{2}} \:−\:\mathrm{yz}\:=\:\mathrm{1}\:\:\:\:\:\:..........\:\left(\mathrm{i}\right) \\ $$$$\mathrm{y}^{\mathrm{2}} \:−\:\mathrm{xz}\:=\:\mathrm{4}\:\:\:\:\:\:..........\:\left(\mathrm{i}\right) \\ $$$$\mathrm{z}^{\mathrm{2}} \:−\:\mathrm{xy}\:=\:\mathrm{9}\:\:\:\:\:\:..........\:\left(\mathrm{i}\right) \\ $$

Question Number 23682 Answers: 0 Comments: 0

Question Number 23644 Answers: 0 Comments: 0

Prove that Σ_(r=0) ^n r.^n C_r z^r = nz(1 + z)^(n−1)

$${Prove}\:{that}\:\underset{{r}=\mathrm{0}} {\overset{{n}} {\sum}}{r}.^{{n}} {C}_{{r}} \:{z}^{{r}} \:=\:{nz}\left(\mathrm{1}\:+\:{z}\right)^{{n}−\mathrm{1}} \\ $$

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

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