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Question Number 66546 by Sayantan chakraborty last updated on 17/Aug/19

Commented by Sayantan chakraborty last updated on 17/Aug/19

urgent need

$$\mathrm{urgent}\:\mathrm{need} \\ $$

Commented by ~ À ® @ 237 ~ last updated on 17/Aug/19

that sum is S_n = Σ_(k=1) ^n ((2^k k)/((k+1)(k+2))) as (n/((n+1)(n+2))) = ((2(n+1)−(n+2))/((n+1)(n+2))) =(2/(n+2)) −(1/(n+1)) now S_n = Σ_(k=1 ) ^n ((2^(k+1) /(k+2)) −(2^k /(k+1)) ) = Σ_(k=2) ^(n+1) (2^k /(k+1)) −Σ_(k=1) ^n (2^k /(k+1))=(2^(n+1) /(n+2)) −1

$$ \\ $$$$ \\ $$$${that}\:{sum}\:{is}\:{S}_{{n}} =\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{2}^{{k}} {k}}{\left({k}+\mathrm{1}\right)\left({k}+\mathrm{2}\right)}\: \\ $$$${as}\:\frac{{n}}{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}\:=\:\frac{\mathrm{2}\left({n}+\mathrm{1}\right)−\left({n}+\mathrm{2}\right)}{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}\:=\frac{\mathrm{2}}{{n}+\mathrm{2}}\:−\frac{\mathrm{1}}{{n}+\mathrm{1}}\: \\ $$$${now}\:\:{S}_{{n}} =\:\underset{{k}=\mathrm{1}\:} {\overset{{n}} {\sum}}\:\left(\frac{\mathrm{2}^{{k}+\mathrm{1}} }{{k}+\mathrm{2}}\:−\frac{\mathrm{2}^{{k}} }{{k}+\mathrm{1}}\:\right)\:=\:\underset{{k}=\mathrm{2}} {\overset{{n}+\mathrm{1}} {\sum}}\frac{\mathrm{2}^{{k}} }{{k}+\mathrm{1}}\:\:\:−\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{2}^{{k}} }{{k}+\mathrm{1}}=\frac{\mathrm{2}^{{n}+\mathrm{1}} }{{n}+\mathrm{2}}\:−\mathrm{1} \\ $$

Commented by Cmr 237 last updated on 17/Aug/19

please i need explication at the end of your demonstration please

$${please}\:{i}\:{need}\:{explication}\:{at}\:{the}\:{end}\:{of}\:\:{your}\:{demonstration}\:{please} \\ $$

Commented by Sayantan chakraborty last updated on 17/Aug/19

what are you want to mean

$$\mathrm{what}\:\mathrm{are}\:\mathrm{you}\:\mathrm{want}\:\mathrm{to}\:\mathrm{mean} \\ $$

Commented by Cmr 237 last updated on 17/Aug/19

thank i understhank

$${thank}\:{i}\:{understhank} \\ $$

Commented by mathmax by abdo last updated on 17/Aug/19

let S_n =(2/(2.3)) +((2^2 .2)/(3.4)) +((2^3 .3)/(4.5)) +....(nterms) ⇒S_n =Σ_(k=1) ^n ((k2^k )/((k+1)(k+2))) S_n =Σ_(k=1) ^n k 2^k {(1/(k+1))−(1/(k+2))} =Σ_(k=1) ^n ((k 2^k )/(k+1))−Σ_(k=1) ^n ((k 2^k )/(k+2)) =Σ_(k=1) ^n (((k+1−1)2^k )/(k+1))−Σ_(k=1) ^n (((k+2−2)2^k )/(k+2)) =Σ_(k=1) ^n 2^k −Σ_(k=1) ^n (2^k /(k+1)) −Σ_(k=1) ^n 2^k +Σ_(k=1) ^n (2^(k+1) /(k+2)) =Σ_(k=3) ^(n+2) (2^(k−1) /k) −Σ_(k=2) ^(n+1) (2^(k−1) /k) =Σ_(k=3) ^(n+1) (2^(k−1) /k) +(2^(n+1) /(n+2))−1−Σ_(k=3) ^(n+1) (2^(k−1) /k) =(2^(n+1) /(n+2))−1 the correct answer is (c)

$${let}\:{S}_{{n}} =\frac{\mathrm{2}}{\mathrm{2}.\mathrm{3}}\:+\frac{\mathrm{2}^{\mathrm{2}} .\mathrm{2}}{\mathrm{3}.\mathrm{4}}\:+\frac{\mathrm{2}^{\mathrm{3}} .\mathrm{3}}{\mathrm{4}.\mathrm{5}}\:+....\left({nterms}\right)\:\Rightarrow{S}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{{k}\mathrm{2}^{{k}} }{\left({k}+\mathrm{1}\right)\left({k}+\mathrm{2}\right)} \\ $$$${S}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}} {k}\:\mathrm{2}^{{k}} \left\{\frac{\mathrm{1}}{{k}+\mathrm{1}}−\frac{\mathrm{1}}{{k}+\mathrm{2}}\right\}\:=\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{{k}\:\mathrm{2}^{{k}} }{{k}+\mathrm{1}}−\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{{k}\:\mathrm{2}^{{k}} }{{k}+\mathrm{2}} \\ $$$$=\sum_{{k}=\mathrm{1}} ^{{n}} \frac{\left({k}+\mathrm{1}−\mathrm{1}\right)\mathrm{2}^{{k}} }{{k}+\mathrm{1}}−\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\left({k}+\mathrm{2}−\mathrm{2}\right)\mathrm{2}^{{k}} }{{k}+\mathrm{2}} \\ $$$$=\sum_{{k}=\mathrm{1}} ^{{n}} \:\mathrm{2}^{{k}} \:−\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{2}^{{k}} }{{k}+\mathrm{1}}\:−\sum_{{k}=\mathrm{1}} ^{{n}} \:\mathrm{2}^{{k}} \:+\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{2}^{{k}+\mathrm{1}} }{{k}+\mathrm{2}} \\ $$$$=\sum_{{k}=\mathrm{3}} ^{{n}+\mathrm{2}} \:\frac{\mathrm{2}^{{k}−\mathrm{1}} }{{k}}\:−\sum_{{k}=\mathrm{2}} ^{{n}+\mathrm{1}} \:\frac{\mathrm{2}^{{k}−\mathrm{1}} }{{k}}\:=\sum_{{k}=\mathrm{3}} ^{{n}+\mathrm{1}} \:\frac{\mathrm{2}^{{k}−\mathrm{1}} }{{k}}\:+\frac{\mathrm{2}^{{n}+\mathrm{1}} }{{n}+\mathrm{2}}−\mathrm{1}−\sum_{{k}=\mathrm{3}} ^{{n}+\mathrm{1}} \:\frac{\mathrm{2}^{{k}−\mathrm{1}} }{{k}} \\ $$$$=\frac{\mathrm{2}^{{n}+\mathrm{1}} }{{n}+\mathrm{2}}−\mathrm{1}\:\:\:\:\:{the}\:{correct}\:{answer}\:{is}\:\left({c}\right) \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\: \\ $$$$ \\ $$

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