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S-k-1-17-k-2-k-




Question Number 103560 by abony1303 last updated on 15/Jul/20
S=Σ_(k=1) ^(17) k∙2^k =?
$$\mathrm{S}=\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{17}} {\sum}}\mathrm{k}\centerdot\mathrm{2}^{\mathrm{k}} =? \\ $$
Commented by abony1303 last updated on 15/Jul/20
pls help
$$\mathrm{pls}\:\mathrm{help} \\ $$
Answered by mr W last updated on 15/Jul/20
S_n =Σ_(k=1) ^n k2^k   2S_n =Σ_(k=1) ^n k2^(k+1)   2S_n =Σ_(k=1) ^n (k+1)2^(k+1) −Σ_(k=1) ^n 2^(k+1)   2S_n =Σ_(k=2) ^(n+1) k2^k −2Σ_(k=1) ^n 2^k   2S_n =Σ_(k=1) ^n k2^k +(n+1)2^(n+1) −2−2×((2(2^n −1))/(2−1))  2S_n =S_n +(n−1)2^(n+1) +2  ⇒S_n =(n−1)2^(n+1) +2
$${S}_{{n}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}\mathrm{2}^{{k}} \\ $$$$\mathrm{2}{S}_{{n}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}\mathrm{2}^{{k}+\mathrm{1}} \\ $$$$\mathrm{2}{S}_{{n}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left({k}+\mathrm{1}\right)\mathrm{2}^{{k}+\mathrm{1}} −\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{2}^{{k}+\mathrm{1}} \\ $$$$\mathrm{2}{S}_{{n}} =\underset{{k}=\mathrm{2}} {\overset{{n}+\mathrm{1}} {\sum}}{k}\mathrm{2}^{{k}} −\mathrm{2}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{2}^{{k}} \\ $$$$\mathrm{2}{S}_{{n}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}\mathrm{2}^{{k}} +\left({n}+\mathrm{1}\right)\mathrm{2}^{{n}+\mathrm{1}} −\mathrm{2}−\mathrm{2}×\frac{\mathrm{2}\left(\mathrm{2}^{{n}} −\mathrm{1}\right)}{\mathrm{2}−\mathrm{1}} \\ $$$$\mathrm{2}{S}_{{n}} ={S}_{{n}} +\left({n}−\mathrm{1}\right)\mathrm{2}^{{n}+\mathrm{1}} +\mathrm{2} \\ $$$$\Rightarrow{S}_{{n}} =\left({n}−\mathrm{1}\right)\mathrm{2}^{{n}+\mathrm{1}} +\mathrm{2} \\ $$
Answered by Dwaipayan Shikari last updated on 15/Jul/20
S_n =1.2+2.2^2 +3.2^3 +...........+n.2^n     2S_n =1.2^2 +2.2^3 +.....+(n−1)2^n +n2^(n+1)   ............Subtracting  −S_n =1.2+1.2^2 +1.2^3 +....+2^n −n2^(n+1)   S_n =n2^(n+1) −(1.2+((2^n −1)/1))  S_n =(n−1)^(2n+1) +2  S_(17) =16.2^(18) +2  (⊛■★  DS)
$${S}_{{n}} =\mathrm{1}.\mathrm{2}+\mathrm{2}.\mathrm{2}^{\mathrm{2}} +\mathrm{3}.\mathrm{2}^{\mathrm{3}} +………..+{n}.\mathrm{2}^{{n}} \\ $$$$ \\ $$$$\mathrm{2}{S}_{{n}} =\mathrm{1}.\mathrm{2}^{\mathrm{2}} +\mathrm{2}.\mathrm{2}^{\mathrm{3}} +…..+\left({n}−\mathrm{1}\right)\mathrm{2}^{{n}} +{n}\mathrm{2}^{{n}+\mathrm{1}} \\ $$$$…………{Subtracting} \\ $$$$−{S}_{{n}} =\mathrm{1}.\mathrm{2}+\mathrm{1}.\mathrm{2}^{\mathrm{2}} +\mathrm{1}.\mathrm{2}^{\mathrm{3}} +….+\mathrm{2}^{{n}} −{n}\mathrm{2}^{{n}+\mathrm{1}} \\ $$$${S}_{{n}} ={n}\mathrm{2}^{{n}+\mathrm{1}} −\left(\mathrm{1}.\mathrm{2}+\frac{\mathrm{2}^{{n}} −\mathrm{1}}{\mathrm{1}}\right) \\ $$$${S}_{{n}} =\left({n}−\mathrm{1}\right)^{\mathrm{2}{n}+\mathrm{1}} +\mathrm{2} \\ $$$${S}_{\mathrm{17}} =\mathrm{16}.\mathrm{2}^{\mathrm{18}} +\mathrm{2}\:\:\left(\circledast\blacksquare\bigstar\:\:\mathrm{DS}\right) \\ $$
Answered by OlafThorendsen last updated on 15/Jul/20
f(x) = Σ_(k=0) ^n x^k  = ((1−x^(n+1) )/(1−x))  f′(x) = Σ_(k=1) ^n kx^(k−1)   and f′(x) = ((−(n+1)x^n (1−x)+(1−x^(n+1) ))/((1−x)^2 ))  xf′(x) = Σ_(k=1) ^n kx^k  = x((1−(n+1)x^n +nx^(n+1) )/((1−x)^2 ))  ⇒Σ_(k=1) ^(17) k2^k  = 2((1−(17+1)2^(17) +17.2^(17+1) )/((1−2)^2 ))  ⇒Σ_(k=1) ^(17) k2^k  = 2(17.2^(18) −18.2^(17) +1)  ⇒Σ_(k=1) ^(17) k2^k  = 4.194.306
$${f}\left({x}\right)\:=\:\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{x}^{{k}} \:=\:\frac{\mathrm{1}−{x}^{{n}+\mathrm{1}} }{\mathrm{1}−{x}} \\ $$$${f}'\left({x}\right)\:=\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{kx}^{{k}−\mathrm{1}} \\ $$$$\mathrm{and}\:{f}'\left({x}\right)\:=\:\frac{−\left({n}+\mathrm{1}\right){x}^{{n}} \left(\mathrm{1}−{x}\right)+\left(\mathrm{1}−{x}^{{n}+\mathrm{1}} \right)}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} } \\ $$$${xf}'\left({x}\right)\:=\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{kx}^{{k}} \:=\:{x}\frac{\mathrm{1}−\left({n}+\mathrm{1}\right){x}^{{n}} +{nx}^{{n}+\mathrm{1}} }{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow\underset{{k}=\mathrm{1}} {\overset{\mathrm{17}} {\sum}}{k}\mathrm{2}^{{k}} \:=\:\mathrm{2}\frac{\mathrm{1}−\left(\mathrm{17}+\mathrm{1}\right)\mathrm{2}^{\mathrm{17}} +\mathrm{17}.\mathrm{2}^{\mathrm{17}+\mathrm{1}} }{\left(\mathrm{1}−\mathrm{2}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow\underset{{k}=\mathrm{1}} {\overset{\mathrm{17}} {\sum}}{k}\mathrm{2}^{{k}} \:=\:\mathrm{2}\left(\mathrm{17}.\mathrm{2}^{\mathrm{18}} −\mathrm{18}.\mathrm{2}^{\mathrm{17}} +\mathrm{1}\right) \\ $$$$\Rightarrow\underset{{k}=\mathrm{1}} {\overset{\mathrm{17}} {\sum}}{k}\mathrm{2}^{{k}} \:=\:\mathrm{4}.\mathrm{194}.\mathrm{306} \\ $$$$ \\ $$

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