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Question Number 74573 by Raxreedoroid last updated on 26/Nov/19

Find (turn it into non-segma expression) 1+Σ_(k=1) ^(n−1) (((−1)^k +3)/2)

$$\mathrm{Find}\:\left(\mathrm{turn}\:\mathrm{it}\:\mathrm{into}\:\mathrm{non}-\mathrm{segma}\:\mathrm{expression}\right) \\ $$$$\mathrm{1}+\underset{{k}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}} +\mathrm{3}}{\mathrm{2}} \\ $$

Commented by mathmax by abdo last updated on 26/Nov/19

let S =1+Σ_(k=1) ^(n−1) ((3+(−1)^k )/2) =1+(3/2)Σ_(k=1) ^(n−1) (1)+(1/2)Σ_(k=1) ^(n−1) (−1)^k =1+(3/2)(n−1) +(1/2)(Σ_(k=0) ^(n−1) (−1)^k −1) =((3n)/2)−(1/2)−(1/2) +(1/2)×((1−(−1)^n )/2) =((3n)/2) −1 +(1/4)(1−(−1)^n ) =((3n)/2)−(3/4) −(((−1)^n )/4)

$${let}\:{S}\:=\mathrm{1}+\sum_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:\frac{\mathrm{3}+\left(−\mathrm{1}\right)^{{k}} }{\mathrm{2}}\:=\mathrm{1}+\frac{\mathrm{3}}{\mathrm{2}}\sum_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \left(\mathrm{1}\right)+\frac{\mathrm{1}}{\mathrm{2}}\sum_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \left(−\mathrm{1}\right)^{{k}} \\ $$$$=\mathrm{1}+\frac{\mathrm{3}}{\mathrm{2}}\left({n}−\mathrm{1}\right)\:+\frac{\mathrm{1}}{\mathrm{2}}\left(\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \left(−\mathrm{1}\right)^{{k}} −\mathrm{1}\right) \\ $$$$=\frac{\mathrm{3}{n}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}−\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}}\:=\frac{\mathrm{3}{n}}{\mathrm{2}}\:−\mathrm{1}\:+\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{1}−\left(−\mathrm{1}\right)^{{n}} \right) \\ $$$$=\frac{\mathrm{3}{n}}{\mathrm{2}}−\frac{\mathrm{3}}{\mathrm{4}}\:−\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{4}} \\ $$

Answered by mind is power last updated on 26/Nov/19

1+Σ_(k=1) ^(n−1) (((−1)^k )/2)+Σ_(k=1) ^(n−1) (3/2) =1+(((−1))/2).((−1−(−1)^(n−1) )/2)+(3/2)(n−1) =(((−1)^(n−1) +6n−3)/4)

$$\mathrm{1}+\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}−\mathrm{1}} {\sum}}\frac{\left(−\mathrm{1}\right)^{\mathrm{k}} }{\mathrm{2}}+\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}−\mathrm{1}} {\sum}}\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$=\mathrm{1}+\frac{\left(−\mathrm{1}\right)}{\mathrm{2}}.\frac{−\mathrm{1}−\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} }{\mathrm{2}}+\frac{\mathrm{3}}{\mathrm{2}}\left(\mathrm{n}−\mathrm{1}\right) \\ $$$$=\frac{\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} +\mathrm{6n}−\mathrm{3}}{\mathrm{4}} \\ $$

Commented by Smail last updated on 26/Nov/19

1−x^n =(1−x)(1+x+x^2 +x^3 +...+x^(n−1) ) ⇔((1−x^n )/(1−x))=1+x+x^2 +...+x^(n−1) Thus, Σ_(k=0) ^(n−1) x^k =((1−x^n )/(1−x))

$$\mathrm{1}−{x}^{{n}} =\left(\mathrm{1}−{x}\right)\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +...+{x}^{{n}−\mathrm{1}} \right) \\ $$$$\Leftrightarrow\frac{\mathrm{1}−{x}^{{n}} }{\mathrm{1}−{x}}=\mathrm{1}+{x}+{x}^{\mathrm{2}} +...+{x}^{{n}−\mathrm{1}} \\ $$$${Thus},\:\:\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}{x}^{{k}} =\frac{\mathrm{1}−{x}^{{n}} }{\mathrm{1}−{x}} \\ $$

Commented by JDamian last updated on 26/Nov/19

Σ_(k=1) ^(n−1) x^k is the sum of a geometric progession

$$\underset{{k}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\sum}}{x}^{{k}} \:\:{is}\:{the}\:{sum}\:{of}\:{a}\:\boldsymbol{{geometric}}\:\boldsymbol{{progession}} \\ $$

Commented by Raxreedoroid last updated on 26/Nov/19

What the step of expanding Σ_(k=1) ^(n−1) (((−1)^k )/2) called? or how it was done?

$$\mathrm{What}\:\mathrm{the}\:\mathrm{step}\:\mathrm{of}\:\mathrm{expanding}\:\underset{{k}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}} }{\mathrm{2}}\:\mathrm{called}?\:\mathrm{or}\:\mathrm{how}\:\mathrm{it}\:\mathrm{was}\:\mathrm{done}? \\ $$

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