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Question Number 90472 by Ar Brandon last updated on 23/Apr/20

Σ_(k=0) ^m ((2k+3)/2^(m−k) )

$$\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{m}} {\sum}}\frac{\mathrm{2k}+\mathrm{3}}{\mathrm{2}^{\mathrm{m}−\mathrm{k}} } \\ $$

Commented by abdomathmax last updated on 23/Apr/20

A_m =(1/2^m )Σ_(k=0) ^m (2 k+3)2^k =(1/2^(m−1) )Σ_(k=0) ^m k 2^k +(3/2^m )Σ_(k=0) ^m 2^k we have Σ_(k=0) ^m x^k =((x^(m+1) −1)/(x−1)) (we suppose x≠1) by derivation we get Σ_(k=1) ^m kx^(k−1) =(((x^(m+1) −1)/(x−1)))^((1)) =((mx^(m+1) −(m+1)x^m +1)/((x−1)^2 )) ⇒ Σ_(k=1) ^m k x^k =((mx^(m+2) −)m+1)x^(m+1) +x)/((x−1)^2 )) ⇒ Σ_(k=0) ^m k2^k =((m2^(m+2) −(m+1)2^(m+1 ) +2)/1^2 ) =4m 2^m −2(m+1)2^m +2 also Σ_(k=0) ^m 2^k =((2^(m+1) −1)/(2−1)) =2^(m+1) −1 ⇒ A_m =(1/2^(m−1) )(4m 2^m −2(m+1)2^m +2) +(3/2^m )(2^(m+1) −1) =8m −4(m+1) +(1/2^(m−2) ) +6−(3/2^m ) =4m +2 +(1/2^m )

$${A}_{{m}} =\frac{\mathrm{1}}{\mathrm{2}^{{m}} }\sum_{{k}=\mathrm{0}} ^{{m}} \left(\mathrm{2}\:{k}+\mathrm{3}\right)\mathrm{2}^{{k}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}^{{m}−\mathrm{1}} }\sum_{{k}=\mathrm{0}} ^{{m}} \:{k}\:\mathrm{2}^{{k}} \:+\frac{\mathrm{3}}{\mathrm{2}^{{m}} }\sum_{{k}=\mathrm{0}} ^{{m}} \:\mathrm{2}^{{k}} \\ $$$${we}\:{have}\:\sum_{{k}=\mathrm{0}} ^{{m}} \:{x}^{{k}} \:=\frac{{x}^{{m}+\mathrm{1}} −\mathrm{1}}{{x}−\mathrm{1}}\:\:\left({we}\:{suppose}\:{x}\neq\mathrm{1}\right)\:{by} \\ $$$${derivation}\:{we}\:{get}\:\sum_{{k}=\mathrm{1}} ^{{m}} \:{kx}^{{k}−\mathrm{1}} \:=\left(\frac{{x}^{{m}+\mathrm{1}} −\mathrm{1}}{{x}−\mathrm{1}}\right)^{\left(\mathrm{1}\right)} \\ $$$$=\frac{{mx}^{{m}+\mathrm{1}} −\left({m}+\mathrm{1}\right){x}^{{m}} \:+\mathrm{1}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$$\sum_{{k}=\mathrm{1}} ^{{m}} \:{k}\:{x}^{{k}} \:=\frac{\left.{m}\left.{x}^{{m}+\mathrm{2}} −\right){m}+\mathrm{1}\right){x}^{{m}+\mathrm{1}} \:+{x}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$$\sum_{{k}=\mathrm{0}} ^{{m}} \:{k}\mathrm{2}^{{k}} \:=\frac{{m}\mathrm{2}^{{m}+\mathrm{2}} −\left({m}+\mathrm{1}\right)\mathrm{2}^{{m}+\mathrm{1}\:} \:+\mathrm{2}}{\mathrm{1}^{\mathrm{2}} } \\ $$$$=\mathrm{4}{m}\:\mathrm{2}^{{m}} −\mathrm{2}\left({m}+\mathrm{1}\right)\mathrm{2}^{{m}} \:+\mathrm{2}\:{also} \\ $$$$\sum_{{k}=\mathrm{0}} ^{{m}} \:\mathrm{2}^{{k}} \:=\frac{\mathrm{2}^{{m}+\mathrm{1}} −\mathrm{1}}{\mathrm{2}−\mathrm{1}}\:=\mathrm{2}^{{m}+\mathrm{1}} −\mathrm{1}\:\Rightarrow \\ $$$${A}_{{m}} =\frac{\mathrm{1}}{\mathrm{2}^{{m}−\mathrm{1}} }\left(\mathrm{4}{m}\:\mathrm{2}^{{m}} −\mathrm{2}\left({m}+\mathrm{1}\right)\mathrm{2}^{{m}} \:+\mathrm{2}\right) \\ $$$$+\frac{\mathrm{3}}{\mathrm{2}^{{m}} }\left(\mathrm{2}^{{m}+\mathrm{1}} −\mathrm{1}\right) \\ $$$$=\mathrm{8}{m}\:−\mathrm{4}\left({m}+\mathrm{1}\right)\:+\frac{\mathrm{1}}{\mathrm{2}^{{m}−\mathrm{2}} }\:+\mathrm{6}−\frac{\mathrm{3}}{\mathrm{2}^{{m}} } \\ $$$$=\mathrm{4}{m}\:+\mathrm{2}\:\:+\frac{\mathrm{1}}{\mathrm{2}^{{m}} } \\ $$

Commented by Ar Brandon last updated on 24/Apr/20

Thanks for the idea.

$$\mathrm{Thanks}\:\mathrm{for}\:\mathrm{the}\:\mathrm{idea}. \\ $$

Commented by mathmax by abdo last updated on 24/Apr/20

you are welcome.

$${you}\:{are}\:{welcome}. \\ $$

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