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Question Number 73044    Answers: 0   Comments: 0

prove that Σ_(k=1) ^n H_k =(n+1)H_n −n and Σ_(k=1) ^n H_k ^2 =(n+1)H_n ^2 −(2n+1)H_n +2n

$${prove}\:{that}\:\:\sum_{{k}=\mathrm{1}} ^{{n}} \:{H}_{{k}} =\left({n}+\mathrm{1}\right){H}_{{n}} −{n} \\ $$$${and}\:\sum_{{k}=\mathrm{1}} ^{{n}} \:{H}_{{k}} ^{\mathrm{2}} \:=\left({n}+\mathrm{1}\right){H}_{{n}} ^{\mathrm{2}} \:−\left(\mathrm{2}{n}+\mathrm{1}\right){H}_{{n}} \:+\mathrm{2}{n} \\ $$

Question Number 73043    Answers: 1   Comments: 2

prove that H_n =Σ_(k=1) ^n (((−1)^(k+1) )/k)×C_n ^k H_n =Σ_(k=1) ^n (1/k)

$${prove}\:{that}\:{H}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\left(−\mathrm{1}\right)^{{k}+\mathrm{1}} }{{k}}×{C}_{{n}} ^{{k}} \\ $$$${H}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}} \\ $$

Question Number 73042    Answers: 1   Comments: 0

prove that for (n,p)∈N^★^2 Σ_(k=0) ^(p ) k C_n ^(p−k) C_n ^k =n C_(2n−1) ^(p−1) conclude the value of Σ_(k=0) ^n k (C_n ^k )^2

$${prove}\:{that}\:{for}\:\left({n},{p}\right)\in{N}^{\bigstar^{\mathrm{2}} } \:\:\:\sum_{{k}=\mathrm{0}} ^{{p}\:} \:{k}\:{C}_{{n}} ^{{p}−{k}} \:{C}_{{n}} ^{{k}} \:={n}\:{C}_{\mathrm{2}{n}−\mathrm{1}} ^{{p}−\mathrm{1}} \\ $$$${conclude}\:{the}\:{value}\:{of}\:\sum_{{k}=\mathrm{0}} ^{{n}} \:{k}\:\left({C}_{{n}} ^{{k}} \right)^{\mathrm{2}} \\ $$

Question Number 73041    Answers: 1   Comments: 1

prove that ∀n∈ N Σ_(k=0) ^(2n) (−1)^k (C_(2n) ^k )^2 =(−1)^n C_(2n) ^n

$${prove}\:{that}\:\forall{n}\in\:{N}\:\:\sum_{{k}=\mathrm{0}} ^{\mathrm{2}{n}} \:\left(−\mathrm{1}\right)^{{k}} \:\left({C}_{\mathrm{2}{n}} ^{{k}} \right)^{\mathrm{2}} \:=\left(−\mathrm{1}\right)^{{n}} \:{C}_{\mathrm{2}{n}} ^{{n}} \\ $$

Question Number 73040    Answers: 1   Comments: 0

prove that ∀(n,p)∈N^★ ×N 1)Σ_(k=0) ^p (−1)^k C_n ^k =(−1)^p C_(n−1) ^p 2)∀(p,q)∈N^2 Σ_(k=0) ^p C_(p+q) ^k C_(p+q−k) ^(p−k) =2^p C_(p+q) ^p

$${prove}\:{that}\:\:\forall\left({n},{p}\right)\in{N}^{\bigstar} ×{N} \\ $$$$\left.\mathrm{1}\right)\sum_{{k}=\mathrm{0}} ^{{p}} \:\left(−\mathrm{1}\right)^{{k}} \:{C}_{{n}} ^{{k}} \:=\left(−\mathrm{1}\right)^{{p}} \:{C}_{{n}−\mathrm{1}} ^{{p}} \\ $$$$\left.\mathrm{2}\right)\forall\left({p},{q}\right)\in{N}^{\mathrm{2}} \:\:\:\:\sum_{{k}=\mathrm{0}} ^{{p}} \:{C}_{{p}+{q}} ^{{k}} \:{C}_{{p}+{q}−{k}} ^{{p}−{k}} \:\:=\mathrm{2}^{{p}} \:{C}_{{p}+{q}} ^{{p}} \\ $$

Question Number 73039    Answers: 1   Comments: 0

let U_n =(n/2) if n even and U_n =((n−1)/2) if n odd let f(n)=Σ_(k=0) ^n U_k prove that ∀(x,y)∈N^2 f(x+y)−f(x−y)=xy

$${let}\:{U}_{{n}} =\frac{{n}}{\mathrm{2}}\:{if}\:{n}\:{even}\:{and}\:{U}_{{n}} =\frac{{n}−\mathrm{1}}{\mathrm{2}}\:{if}\:{n}\:{odd}\:{let}\:{f}\left({n}\right)=\sum_{{k}=\mathrm{0}} ^{{n}} {U}_{{k}} \\ $$$${prove}\:{that}\:\forall\left({x},{y}\right)\in{N}^{\mathrm{2}} \:\:\:\:{f}\left({x}+{y}\right)−{f}\left({x}−{y}\right)={xy} \\ $$

Question Number 73037    Answers: 1   Comments: 1

Question Number 73036    Answers: 1   Comments: 3

calculate 1) Σ_(k=1) ^n k^2 (n+1−k) 2)Σ_(1≤i≤j≤n) ij

$$\left.{calculate}\:\mathrm{1}\right)\:\sum_{{k}=\mathrm{1}} ^{{n}} \:{k}^{\mathrm{2}} \left({n}+\mathrm{1}−{k}\right) \\ $$$$\left.\mathrm{2}\right)\sum_{\mathrm{1}\leqslant{i}\leqslant{j}\leqslant{n}} \:{ij} \\ $$

Question Number 73035    Answers: 0   Comments: 0

prove that ∀n∈N^★ 2!4!....(2n)!≥{(n+1)!}^n

$${prove}\:{that}\:\:\forall{n}\in{N}^{\bigstar} \:\:\:\:\:\mathrm{2}!\mathrm{4}!....\left(\mathrm{2}{n}\right)!\geqslant\left\{\left({n}+\mathrm{1}\right)!\right\}^{{n}} \\ $$

Question Number 73034    Answers: 1   Comments: 1

calculate U_n =Σ_(k=1) ^n (k/((k+1)!))

$${calculate}\:{U}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{{k}}{\left({k}+\mathrm{1}\right)!} \\ $$

Question Number 73033    Answers: 1   Comments: 0

solve inside N^2 x(x+1)=4y(y+1)

$${solve}\:{inside}\:{N}^{\mathrm{2}} \:\:\:\:{x}\left({x}+\mathrm{1}\right)=\mathrm{4}{y}\left({y}+\mathrm{1}\right) \\ $$

Question Number 73032    Answers: 1   Comments: 0

find x from n / ∃n∈N^n and 1+x+x^2 +x^3 +x^4 =n^2

$${find}\:{x}\:{from}\:{n}\:\:/\:\exists{n}\in{N}^{{n}} \:\:\:\:{and}\:\mathrm{1}+{x}+{x}^{\mathrm{2}} \:+{x}^{\mathrm{3}} \:+{x}^{\mathrm{4}} ={n}^{\mathrm{2}} \\ $$

Question Number 73031    Answers: 0   Comments: 0

solve inside N^2 3x^3 +xy +4y^3 =349

$${solve}\:{inside}\:{N}^{\mathrm{2}} \:\:\:\mathrm{3}{x}^{\mathrm{3}} \:+{xy}\:+\mathrm{4}{y}^{\mathrm{3}} \:=\mathrm{349} \\ $$

Question Number 73030    Answers: 2   Comments: 0

Question Number 73029    Answers: 1   Comments: 0

prove that ∀(n,p,q)∈N^3 Σ_(k=0) ^n C_p ^k C_q ^(n−k) =C_(p+q) ^n conclude that Σ_(k=0) ^n (C_n ^k )^2 =C_(2n) ^n

$${prove}\:{that}\:\:\forall\left({n},{p},{q}\right)\in{N}^{\mathrm{3}} \:\:\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{p}} ^{{k}} \:{C}_{{q}} ^{{n}−{k}} \:\:\:={C}_{{p}+{q}} ^{{n}} \\ $$$${conclude}\:{that}\:\sum_{{k}=\mathrm{0}} ^{{n}} \:\left({C}_{{n}} ^{{k}} \right)^{\mathrm{2}} \:={C}_{\mathrm{2}{n}} ^{{n}} \\ $$

Question Number 73028    Answers: 2   Comments: 0

calculate Σ_(1≤i≤n and 1≤j≤n) min(i,j)

$${calculate}\:\sum_{\mathrm{1}\leqslant{i}\leqslant{n}\:{and}\:\mathrm{1}\leqslant{j}\leqslant{n}} \:\:{min}\left({i},{j}\right) \\ $$

Question Number 73027    Answers: 1   Comments: 0

x and y are reals(or complex) let put x^((0)) =1 ,x^((1)) =x x^((2)) =x(x−1).....x^((n)) =x(x−1)(x−2)...(x−n+1)prove that (x+y)^((n)) =Σ_(k=0) ^n C_n ^k x^((n−k)) y^((k))

$${x}\:{and}\:{y}\:{are}\:{reals}\left({or}\:{complex}\right)\:{let}\:{put}\:{x}^{\left(\mathrm{0}\right)} =\mathrm{1}\:,{x}^{\left(\mathrm{1}\right)} ={x} \\ $$$${x}^{\left(\mathrm{2}\right)} ={x}\left({x}−\mathrm{1}\right).....{x}^{\left({n}\right)} ={x}\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)...\left({x}−{n}+\mathrm{1}\right){prove}\:{that} \\ $$$$\left({x}+{y}\right)^{\left({n}\right)} =\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\:{x}^{\left({n}−{k}\right)} {y}^{\left({k}\right)} \\ $$

Question Number 73021    Answers: 0   Comments: 0

Question Number 73017    Answers: 0   Comments: 1

find lim_(x→+∞) x(√(x^2 + 1))

$${find}\: \\ $$$$\:\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\:\:{x}\sqrt{{x}^{\mathrm{2}} \:+\:\mathrm{1}}\: \\ $$

Question Number 73084    Answers: 1   Comments: 0

Question Number 72998    Answers: 0   Comments: 0

The acute angle of the rectangle trapezius is equal to α=90°arcsin0.1 The bases measure 10 and 30. Calculate the area of the trapezius.

$${The}\:{acute}\:{angle}\:{of}\:{the}\:{rectangle}\:{trapezius}\:{is}\:{equal}\:{to}\:\alpha=\mathrm{90}°{arcsin}\mathrm{0}.\mathrm{1} \\ $$$${The}\:{bases}\:{measure}\:\mathrm{10}\:{and}\:\mathrm{30}.\:{Calculate}\:{the}\:{area}\:{of}\:{the}\:{trapezius}. \\ $$

Question Number 72997    Answers: 1   Comments: 0

The area of the equilateral triangle is equal to (((√(16))(√8))/(3(√π))) Calculate the area of the circle inscribed in the triangle.

$${The}\:{area}\:{of}\:{the}\:{equilateral}\:{triangle}\:{is}\:{equal}\:{to}\:\frac{\sqrt{\mathrm{16}}\sqrt{\mathrm{8}}}{\mathrm{3}\sqrt{\pi}} \\ $$$${Calculate}\:{the}\:{area}\:{of}\:{the}\:{circle}\:{inscribed}\:{in}\:{the}\:{triangle}. \\ $$$$\: \\ $$

Question Number 72990    Answers: 1   Comments: 1

calculate lim_(x→0) ((arctan(e^x )−(π/4))/x^2 )

$${calculate}\:{lim}_{{x}\rightarrow\mathrm{0}} \:\:\:\:\:\frac{{arctan}\left({e}^{{x}} \right)−\frac{\pi}{\mathrm{4}}}{{x}^{\mathrm{2}} } \\ $$

Question Number 72988    Answers: 1   Comments: 1

calculate f(x)=∫_0 ^∞ (e^(−xt^2 ) /(4+t^2 ))dt with x>0

$${calculate}\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{e}^{−{xt}^{\mathrm{2}} } }{\mathrm{4}+{t}^{\mathrm{2}} }{dt}\:\:\:{with}\:{x}>\mathrm{0} \\ $$

Question Number 72986    Answers: 0   Comments: 0

calculate ∫_0 ^∞ ((e^(−x^2 ) cosx)/((x^2 +x+1)^2 ))dx

$${calculate}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{e}^{−{x}^{\mathrm{2}} } \:\:{cosx}}{\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)^{\mathrm{2}} }{dx} \\ $$

Question Number 72965    Answers: 1   Comments: 1

Prove that this equation is true: Π_(x=1) ^(n−1) (2x+1)=(((2x−1)!)/((2)^(x−1) (x−1)!))

$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{this}\:\mathrm{equation}\:\mathrm{is}\:\mathrm{true}: \\ $$$$\underset{{x}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\prod}}\left(\mathrm{2}{x}+\mathrm{1}\right)=\frac{\left(\mathrm{2}{x}−\mathrm{1}\right)!}{\left(\mathrm{2}\right)^{{x}−\mathrm{1}} \left({x}−\mathrm{1}\right)!} \\ $$

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