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let k be natural number. defined s_k as the sum of the infinite series s_k =((k^2 −1)/k^0 ) + ((k^2 −1)/k^1 ) + ((k^2 −1)/k^2 ) +... find the value of Σ_(k=1) ^∞ [(s_k /2^(k−1) )] .

$${let}\:{k}\:{be}\:{natural}\:{number}.\:{defined}\:{s}_{{k}} \:{as}\:{the} \\ $$$${sum}\:{of}\:{the}\:{infinite}\:{series}\:{s}_{{k}} =\frac{{k}^{\mathrm{2}} −\mathrm{1}}{{k}^{\mathrm{0}} }\:+\:\frac{{k}^{\mathrm{2}} −\mathrm{1}}{{k}^{\mathrm{1}} }\:+\:\frac{{k}^{\mathrm{2}} −\mathrm{1}}{{k}^{\mathrm{2}} }\:+... \\ $$$${find}\:{the}\:{value}\:{of}\:\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\:\left[\frac{{s}_{{k}} }{\mathrm{2}^{{k}−\mathrm{1}} }\right]\:\:. \\ $$

Question Number 192606 Answers: 0 Comments: 1

proof that 0^0 =1 without limit!

$${proof}\:{that}\:\mathrm{0}^{\mathrm{0}} =\mathrm{1}\:{without}\:{limit}! \\ $$

Question Number 192604 Answers: 0 Comments: 1

many people say that the 0^(0 ) is an uninfinity ones of them say that 0^0 is infinity and equal to 1! what do you think wich ones of them say right?

$${many}\:{people}\:{say}\:{that}\:{the}\:\mathrm{0}^{\mathrm{0}\:} {is}\:{an}\:{uninfinity} \\ $$$${ones}\:{of}\:{them}\:{say}\:{that}\:\mathrm{0}^{\mathrm{0}} \:{is}\:{infinity}\:{and}\:{equal}\:{to}\:\mathrm{1}! \\ $$$${what}\:{do}\:{you}\:{think}\:{wich}\:{ones}\:{of}\:{them} \\ $$$${say}\:{right}? \\ $$

Question Number 192597 Answers: 1 Comments: 0

a_(1 ) , a_2 , a_(3 ) , .... , a_n is a sequence satifies that a_(n+2) =a_(n+1) −a_n for n ≥ 1. suppose the sum of the first 999 terms = 1003 and the sum of the first 1003 terms = −999 find the sum of the first 2002 terms.

$${a}_{\mathrm{1}\:} \:,\:{a}_{\mathrm{2}} \:,\:{a}_{\mathrm{3}\:} \:,\:....\:,\:{a}_{{n}} \:{is}\:{a}\:{sequence}\:{satifies}\:{that} \\ $$$${a}_{{n}+\mathrm{2}} ={a}_{{n}+\mathrm{1}} −{a}_{{n}} \:{for}\:{n}\:\geq\:\mathrm{1}.\:{suppose}\:{the}\:{sum}\: \\ $$$${of}\:{the}\:{first}\:\mathrm{999}\:{terms}\:=\:\mathrm{1003}\:{and}\:{the}\:{sum} \\ $$$${of}\:{the}\:{first}\:\mathrm{1003}\:{terms}\:=\:−\mathrm{999}\:{find}\:{the}\: \\ $$$${sum}\:{of}\:{the}\:{first}\:\mathrm{2002}\:{terms}. \\ $$

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Question Number 192590 Answers: 1 Comments: 0

Σ_(n=1) ^∞ Σ_(m=1) ^∞ [((m^2 n)/(3^m (3^n .m+3^m .n)))]=λ find λ.

$$ \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\underset{{m}=\mathrm{1}} {\overset{\infty} {\sum}}\left[\frac{{m}^{\mathrm{2}} {n}}{\mathrm{3}^{{m}} \left(\mathrm{3}^{{n}} .{m}+\mathrm{3}^{{m}} .{n}\right)}\right]=\lambda\: \\ $$$${find}\:\lambda. \\ $$$$ \\ $$

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This might be helpful: How to easily calculate z_1 ^z_2 with z_1 , z_2 ∈C: Transform z_1 =re^(iθ) and z_2 =a+bi ⇒ z_1 ^z_2 =(re^(iθ) )^(a+bi) z_1 ^z_2 =r^(a+bi) e^(−bθ+iaθ) z_1 ^z_2 =r^a e^(−bθ) r^(bi) e^(iaθ) z_1 ^z_2 =(r^a /e^(bθ) )e^(i(aθ+bln r))

$$\mathrm{This}\:\mathrm{might}\:\mathrm{be}\:\mathrm{helpful}: \\ $$$$\mathrm{How}\:\mathrm{to}\:\mathrm{easily}\:\mathrm{calculate}\:{z}_{\mathrm{1}} ^{{z}_{\mathrm{2}} } \:\mathrm{with}\:{z}_{\mathrm{1}} ,\:{z}_{\mathrm{2}} \:\in\mathbb{C}: \\ $$$$\mathrm{Transform}\:{z}_{\mathrm{1}} ={r}\mathrm{e}^{\mathrm{i}\theta} \:\mathrm{and}\:{z}_{\mathrm{2}} ={a}+{b}\mathrm{i}\:\Rightarrow \\ $$$${z}_{\mathrm{1}} ^{{z}_{\mathrm{2}} } =\left({r}\mathrm{e}^{\mathrm{i}\theta} \right)^{{a}+{b}\mathrm{i}} \\ $$$${z}_{\mathrm{1}} ^{{z}_{\mathrm{2}} } ={r}^{{a}+{b}\mathrm{i}} \mathrm{e}^{−{b}\theta+\mathrm{i}{a}\theta} \\ $$$${z}_{\mathrm{1}} ^{{z}_{\mathrm{2}} } ={r}^{{a}} \mathrm{e}^{−{b}\theta} {r}^{{b}\mathrm{i}} \mathrm{e}^{\mathrm{i}{a}\theta} \\ $$$${z}_{\mathrm{1}} ^{{z}_{\mathrm{2}} } =\frac{{r}^{{a}} }{\mathrm{e}^{{b}\theta} }\mathrm{e}^{\mathrm{i}\left({a}\theta+{b}\mathrm{ln}\:{r}\right)} \\ $$

Question Number 192560 Answers: 1 Comments: 3

Question Number 192572 Answers: 0 Comments: 0

Let ABCD be a rectangle having an area of 290. Let E be on BC such that BE : BC = 3 : 2. Let F be on CD such that CF : FD = 3 : 1. If G is the intersection of AE and BF, compute the area of △BEG.

$$\mathrm{Let}\:\:{ABCD}\:\:\mathrm{be}\:\:\mathrm{a}\:\:\mathrm{rectangle}\:\:\mathrm{having}\:\:\mathrm{an}\:\mathrm{area}\:\:\mathrm{of}\:\:\mathrm{290}. \\ $$$$\mathrm{Let}\:\:{E}\:\:\mathrm{be}\:\:\mathrm{on}\:\:{BC}\:\:\mathrm{such}\:\:\mathrm{that}\:\:{BE}\::\:{BC}\:=\:\mathrm{3}\::\:\mathrm{2}. \\ $$$$\mathrm{Let}\:\:{F}\:\:\mathrm{be}\:\:\mathrm{on}\:\:{CD}\:\:\mathrm{such}\:\:\mathrm{that}\:\:{CF}\::\:{FD}\:=\:\mathrm{3}\::\:\mathrm{1}. \\ $$$$\mathrm{If}\:\:{G}\:\:\mathrm{is}\:\:\mathrm{the}\:\:\mathrm{intersection}\:\:\mathrm{of}\:\:{AE}\:\:\mathrm{and}\:\:{BF},\:\:\mathrm{compute} \\ $$$$\mathrm{the}\:\:\mathrm{area}\:\:\mathrm{of}\:\:\bigtriangleup{BEG}. \\ $$

Question Number 192570 Answers: 2 Comments: 0

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