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Question Number 40370 by math khazana by abdo last updated on 20/Jul/18
let u_n =Σ_(k=0) ^n (3k+1)(−1)^k 1) calculate interms of n S_n =u_0 +u_1 +u_2 +....+u_n 2) calculate u_0 +u_1 +u_2 +....+u_(57)
$${let}\:{u}_{{n}} \:=\sum_{{k}=\mathrm{0}} ^{{n}} \left(\mathrm{3}{k}+\mathrm{1}\right)\left(−\mathrm{1}\right)^{{k}} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{interms}\:{of}\:{n} \\ $$$${S}_{{n}} ={u}_{\mathrm{0}} \:+{u}_{\mathrm{1}} +{u}_{\mathrm{2}} +….+{u}_{{n}} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:{u}_{\mathrm{0}} \:+{u}_{\mathrm{1}} +{u}_{\mathrm{2}} +….+{u}_{\mathrm{57}} \\ $$
Commented by prof Abdo imad last updated on 20/Jul/18
1)we have u_n =3Σ_(k=0) ^n k(−1)^(k ) +Σ_(k=0) ^n (−1)^k but Σ_(k=0) ^n (−1)^k =((1−(−1)^(n+1) )/2) =((1+(−1)^n )/2) let p(x)=Σ_(k=0) ^n x^k with x≠1 we have p^′ (x)=Σ_(k=1) ^n k x^(k−1) ⇒x p^′ (x)=Σ_(k=1) ^n k x^k ⇒ Σ_(k=1) ^n k(−1)^k =−p^′ (−1) but p(x)=((x^(n+1) −1)/(x−1)) ⇒ p^′ (x) =((nx^(n+1) −(n+1)x^n +1)/((x−1)^2 )) ⇒ p^′ (−1) =((n(−1)^(n+1) −(n+1)(−1)^n +1)/4) ⇒ u_n =−(3/4){ 1−(2n+1)(−1)^n } +((1+(−1)^n )/2) u_n =(3/4){(2n+1)(−1)^n −1} +((1+(−1)^n )/2) we have S_n =Σ_(k=0) ^n u_k =(3/4) Σ_(k=0) ^n (2k+1)(−1)^k −(3/4)Σ_(k=0) ^n (1) +(1/2)Σ_(k=0) ^n (1) +(1/2)Σ_(k=0) ^n (−1)^k =(3/2)Σ_(k=0) ^n k(−1)^k +(3/4)Σ_(k=0) ^n (−1)^k −(3/4)(n+1) +((n+1)/2) +(1/2) ((1+(−1)^n )/2) =(3/2){((1−(2n+1)(−1)^n )/4)} +(3/4) ((1+(−1)^n )/2) −(1/4)(n+1) +(1/4){1+(−1)^n } S_n =(3/8){1−(2n+1)(−1)^n } +(5/8){1+(−1)^n } −((n+1)/4) .
$$\left.\mathrm{1}\right){we}\:{have}\:\:{u}_{{n}} =\mathrm{3}\sum_{{k}=\mathrm{0}} ^{{n}} {k}\left(−\mathrm{1}\right)^{{k}\:} \:+\sum_{{k}=\mathrm{0}} ^{{n}} \left(−\mathrm{1}\right)^{{k}} \:{but} \\ $$$$\sum_{{k}=\mathrm{0}} ^{{n}} \left(−\mathrm{1}\right)^{{k}} =\frac{\mathrm{1}−\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{\mathrm{2}}\:=\frac{\mathrm{1}+\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}} \\ $$$${let}\:{p}\left({x}\right)=\sum_{{k}=\mathrm{0}} ^{{n}} \:{x}^{{k}} \:\:\:\:{with}\:{x}\neq\mathrm{1}\:\:{we}\:{have} \\ $$$${p}^{'} \left({x}\right)=\sum_{{k}=\mathrm{1}} ^{{n}} {k}\:{x}^{{k}−\mathrm{1}} \:\Rightarrow{x}\:{p}^{'} \left({x}\right)=\sum_{{k}=\mathrm{1}} ^{{n}} {k}\:{x}^{{k}} \Rightarrow \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} {k}\left(−\mathrm{1}\right)^{{k}} =−{p}^{'} \left(−\mathrm{1}\right)\:{but}\:{p}\left({x}\right)=\frac{{x}^{{n}+\mathrm{1}} −\mathrm{1}}{{x}−\mathrm{1}}\:\Rightarrow \\ $$$${p}^{'} \left({x}\right)\:=\frac{{nx}^{{n}+\mathrm{1}} −\left({n}+\mathrm{1}\right){x}^{{n}} +\mathrm{1}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$${p}^{'} \left(−\mathrm{1}\right)\:=\frac{{n}\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} −\left({n}+\mathrm{1}\right)\left(−\mathrm{1}\right)^{{n}} \:+\mathrm{1}}{\mathrm{4}}\:\Rightarrow \\ $$$${u}_{{n}} \:=−\frac{\mathrm{3}}{\mathrm{4}}\left\{\:\mathrm{1}−\left(\mathrm{2}{n}+\mathrm{1}\right)\left(−\mathrm{1}\right)^{{n}} \right\}\:+\frac{\mathrm{1}+\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}} \\ $$$${u}_{{n}} =\frac{\mathrm{3}}{\mathrm{4}}\left\{\left(\mathrm{2}{n}+\mathrm{1}\right)\left(−\mathrm{1}\right)^{{n}} −\mathrm{1}\right\}\:+\frac{\mathrm{1}+\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}}\:{we}\:{have} \\ $$$${S}_{{n}} =\sum_{{k}=\mathrm{0}} ^{{n}} \:{u}_{{k}} \\ $$$$=\frac{\mathrm{3}}{\mathrm{4}}\:\sum_{{k}=\mathrm{0}} ^{{n}} \left(\mathrm{2}{k}+\mathrm{1}\right)\left(−\mathrm{1}\right)^{{k}} \:−\frac{\mathrm{3}}{\mathrm{4}}\sum_{{k}=\mathrm{0}} ^{{n}} \left(\mathrm{1}\right) \\ $$$$+\frac{\mathrm{1}}{\mathrm{2}}\sum_{{k}=\mathrm{0}} ^{{n}} \left(\mathrm{1}\right)\:+\frac{\mathrm{1}}{\mathrm{2}}\sum_{{k}=\mathrm{0}} ^{{n}} \left(−\mathrm{1}\right)^{{k}} \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}\sum_{{k}=\mathrm{0}} ^{{n}} {k}\left(−\mathrm{1}\right)^{{k}} \:+\frac{\mathrm{3}}{\mathrm{4}}\sum_{{k}=\mathrm{0}} ^{{n}} \left(−\mathrm{1}\right)^{{k}} −\frac{\mathrm{3}}{\mathrm{4}}\left({n}+\mathrm{1}\right) \\ $$$$+\frac{{n}+\mathrm{1}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{2}}\:\frac{\mathrm{1}+\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}} \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}\left\{\frac{\mathrm{1}−\left(\mathrm{2}{n}+\mathrm{1}\right)\left(−\mathrm{1}\right)^{{n}} }{\mathrm{4}}\right\}\:+\frac{\mathrm{3}}{\mathrm{4}}\:\frac{\mathrm{1}+\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}} \\ $$$$−\frac{\mathrm{1}}{\mathrm{4}}\left({n}+\mathrm{1}\right)\:+\frac{\mathrm{1}}{\mathrm{4}}\left\{\mathrm{1}+\left(−\mathrm{1}\right)^{{n}} \right\} \\ $$$${S}_{{n}} \:=\frac{\mathrm{3}}{\mathrm{8}}\left\{\mathrm{1}−\left(\mathrm{2}{n}+\mathrm{1}\right)\left(−\mathrm{1}\right)^{{n}} \right\}\:+\frac{\mathrm{5}}{\mathrm{8}}\left\{\mathrm{1}+\left(−\mathrm{1}\right)^{{n}} \right\} \\ $$$$−\frac{{n}+\mathrm{1}}{\mathrm{4}}\:. \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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