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The value of Σ_(n=0) ^∞ (((3_n )(2_n )x^n )/((1_n )n!)) β(2,n+1) is a. (1/2)Σ_(n=0) ^∞ (2_n )(x^n /(n!)) b. (1/2)Σ_(n=0) ^∞ (((3_n )(2_n ))/((1_n ))) (x^n /(n!)) c. (1/2)Σ_(n=0) ^∞ (((2_n )x^n )/((1_n )n!)) d. (1/3)Σ_(n=0) ^∞ (((3_n )x^n )/((1_n )n!))

$${The}\:{value}\:{of}\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\left(\mathrm{3}_{{n}} \right)\left(\mathrm{2}_{{n}} \right){x}^{{n}} }{\left(\mathrm{1}_{{n}} \right){n}!}\:\beta\left(\mathrm{2},{n}+\mathrm{1}\right)\:{is} \\ $$$${a}.\:\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\mathrm{2}_{{n}} \right)\frac{{x}^{{n}} }{{n}!} \\ $$$${b}.\:\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\mathrm{3}_{{n}} \right)\left(\mathrm{2}_{{n}} \right)}{\left(\mathrm{1}_{{n}} \right)}\:\frac{{x}^{{n}} }{{n}!} \\ $$$${c}.\:\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\mathrm{2}_{{n}} \right){x}^{{n}} }{\left(\mathrm{1}_{{n}} \right){n}!} \\ $$$${d}.\:\frac{\mathrm{1}}{\mathrm{3}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\mathrm{3}_{{n}} \right){x}^{{n}} }{\left(\mathrm{1}_{{n}} \right){n}!} \\ $$

Question Number 153458 Answers: 0 Comments: 1

Given a set consisting of 22 integer A={±a_1 ,±a_2 ,...,±a_(11) }. Show that exist subset of S with properties (1) for every i=1,2,3,...,11 have least one between a_i or −a_i element of S (2)the sum all possible numbers in S divisible by 2015

$${Given}\:{a}\:{set}\:{consisting}\:{of}\:\mathrm{22}\:{integer} \\ $$$$\:{A}=\left\{\pm{a}_{\mathrm{1}} ,\pm{a}_{\mathrm{2}} ,...,\pm{a}_{\mathrm{11}} \right\}.\:{Show}\:{that} \\ $$$${exist}\:{subset}\:{of}\:{S}\:{with}\:{properties} \\ $$$$\left(\mathrm{1}\right)\:{for}\:{every}\:{i}=\mathrm{1},\mathrm{2},\mathrm{3},...,\mathrm{11}\: \\ $$$$\:{have}\:{least}\:{one}\:{between}\:{a}_{{i}} \:{or}\:−{a}_{{i}} \\ $$$$\:{element}\:{of}\:{S} \\ $$$$\left(\mathrm{2}\right){the}\:{sum}\:{all}\:{possible}\:{numbers} \\ $$$${in}\:{S}\:{divisible}\:{by}\:\mathrm{2015} \\ $$

Question Number 151265 Answers: 1 Comments: 3

Question Number 150984 Answers: 0 Comments: 0

Question Number 150965 Answers: 1 Comments: 0

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Question Number 150223 Answers: 2 Comments: 0

Question Number 149962 Answers: 0 Comments: 0

⌊x⌋+⌊y⌋=43.8 and x+y−⌊x⌋=18.4 .Find 100(x+y).

$$\:\lfloor{x}\rfloor+\lfloor{y}\rfloor=\mathrm{43}.\mathrm{8}\:{and}\:{x}+{y}−\lfloor{x}\rfloor=\mathrm{18}.\mathrm{4} \\ $$$$.{Find}\:\mathrm{100}\left({x}+{y}\right). \\ $$

Question Number 148951 Answers: 2 Comments: 0

Let complex number z=(a+cos θ)+(2a−sin θ)i . If ∣z∣ ≤2 for any θ∈R then the range of real number a is ___

$${Let}\:{complex}\:{number}\:{z}=\left({a}+\mathrm{cos}\:\theta\right)+\left(\mathrm{2}{a}−\mathrm{sin}\:\theta\right){i}\:. \\ $$$${If}\:\mid{z}\mid\:\leqslant\mathrm{2}\:{for}\:{any}\:\theta\in{R}\:{then}\:{the} \\ $$$${range}\:{of}\:{real}\:{number}\:{a}\:{is}\:\_\_\_ \\ $$

Question Number 148211 Answers: 1 Comments: 0

Question Number 145408 Answers: 0 Comments: 0

∫_0 ^x ⌊u⌋(⌊u⌋+1)f(u)du=Σ_(n=1) ^(⌊x⌋) n∫_n ^x f(u)du Prove that

$$\int_{\mathrm{0}} ^{{x}} \lfloor{u}\rfloor\left(\lfloor{u}\rfloor+\mathrm{1}\right){f}\left({u}\right){du}=\underset{{n}=\mathrm{1}} {\overset{\lfloor{x}\rfloor} {\sum}}{n}\int_{{n}} ^{{x}} {f}\left({u}\right){du}\:\: \\ $$$${Prove}\:{that} \\ $$

Question Number 145359 Answers: 1 Comments: 0

How many digits will there be in 875^(16) ?

$$\mathrm{How}\:\mathrm{many}\:\mathrm{digits}\:\mathrm{will}\:\mathrm{there}\:\mathrm{be} \\ $$$$\mathrm{in}\:\mathrm{875}^{\mathrm{16}} \:? \\ $$

Question Number 143790 Answers: 0 Comments: 2

Π_(n=1) ^∞ (1+(x^3 /n^3 ))

$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\left(\mathrm{1}+\frac{{x}^{\mathrm{3}} }{{n}^{\mathrm{3}} }\right) \\ $$

Question Number 143085 Answers: 0 Comments: 0

φ(n^4 +1)=8n φ:Euler totient function Solve for n∈N

$$\phi\left({n}^{\mathrm{4}} +\mathrm{1}\right)=\mathrm{8}{n}\:\:\:\:\:\:\phi:{Euler}\:{totient}\:{function} \\ $$$${Solve}\:{for}\:{n}\in\mathbb{N} \\ $$

Question Number 142880 Answers: 1 Comments: 0

Prove that 𝛗(n)=nΠ_k (1−(1/p_k )) φ(n):Euler totient function

$$\:{Prove}\:{that}\:\boldsymbol{\phi}\left({n}\right)={n}\underset{{k}} {\prod}\left(\mathrm{1}−\frac{\mathrm{1}}{{p}_{{k}} }\right)\:\:\phi\left({n}\right):{Euler}\:{totient}\:{function} \\ $$

Question Number 142263 Answers: 0 Comments: 0

(((0 sin(x))),((0 0)) )!+ (((0 sin(2x))),((0 0)) )!+ (((0 sin(3x))),((0 0)) )!+... n^(th) term

$$\begin{pmatrix}{\mathrm{0}\:{sin}\left({x}\right)}\\{\mathrm{0}\:\:\mathrm{0}}\end{pmatrix}!+\begin{pmatrix}{\mathrm{0}\:\:{sin}\left(\mathrm{2}{x}\right)}\\{\mathrm{0}\:\:\:\:\:\:\:\:\mathrm{0}}\end{pmatrix}!+\begin{pmatrix}{\mathrm{0}\:\:\:{sin}\left(\mathrm{3}{x}\right)}\\{\mathrm{0}\:\:\:\:\:\:\:\:\mathrm{0}}\end{pmatrix}!+...\:{n}^{{th}} \:{term} \\ $$

Question Number 141694 Answers: 0 Comments: 0

log((((√5)+1)/(10))9e^γ )=((ζ(2))/2)(((1^2 +9^2 )/(10^2 )))−((ζ(3))/3) (((1^3 +9^3 )/(10^3 )) )+((ζ(4))/4)(((1^4 +9^4 )/(10^4 )))−... γ=Euler Mascheroni Constant

$${log}\left(\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{10}}\mathrm{9}{e}^{\gamma} \right)=\frac{\zeta\left(\mathrm{2}\right)}{\mathrm{2}}\left(\frac{\mathrm{1}^{\mathrm{2}} +\mathrm{9}^{\mathrm{2}} }{\mathrm{10}^{\mathrm{2}} }\right)−\frac{\zeta\left(\mathrm{3}\right)}{\mathrm{3}}\:\left(\frac{\mathrm{1}^{\mathrm{3}} +\mathrm{9}^{\mathrm{3}} }{\mathrm{10}^{\mathrm{3}} }\:\right)+\frac{\zeta\left(\mathrm{4}\right)}{\mathrm{4}}\left(\frac{\mathrm{1}^{\mathrm{4}} +\mathrm{9}^{\mathrm{4}} }{\mathrm{10}^{\mathrm{4}} }\right)−... \\ $$$$\gamma={Euler}\:{Mascheroni}\:{Constant} \\ $$

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