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Question-120529




Question Number 120529 by bemath last updated on 01/Nov/20
Commented by bemath last updated on 01/Nov/20
Answered by Dwaipayan Shikari last updated on 01/Nov/20
       S=1+2a+3a^2 +4a^3 +.....+na^(n−1)   −aS=     −a−2a^2 −3a^3 −......(n−1)a^(n−1) −na^n   S(1−a)=1+a+a^2 +a^3 +...a^(n−1) −na^n   S(1−a)=((1−a^n )/(1−a))−na^n   S=((1−a^n )/((1−a)^2 ))−na^n  =((na^n )/(a−1))−((a^n −1)/((a−1)^2 ))  n=50  S=50.2^(50) −2^(50) +1=49.2^(50) +1
$$\:\:\:\:\:\:\:{S}=\mathrm{1}+\mathrm{2}{a}+\mathrm{3}{a}^{\mathrm{2}} +\mathrm{4}{a}^{\mathrm{3}} +…..+{na}^{{n}−\mathrm{1}} \\ $$$$−{aS}=\:\:\:\:\:−{a}−\mathrm{2}{a}^{\mathrm{2}} −\mathrm{3}{a}^{\mathrm{3}} −……\left({n}−\mathrm{1}\right){a}^{{n}−\mathrm{1}} −{na}^{{n}} \\ $$$${S}\left(\mathrm{1}−{a}\right)=\mathrm{1}+{a}+{a}^{\mathrm{2}} +{a}^{\mathrm{3}} +…{a}^{{n}−\mathrm{1}} −{na}^{{n}} \\ $$$${S}\left(\mathrm{1}−{a}\right)=\frac{\mathrm{1}−{a}^{{n}} }{\mathrm{1}−{a}}−{na}^{{n}} \\ $$$${S}=\frac{\mathrm{1}−{a}^{{n}} }{\left(\mathrm{1}−{a}\right)^{\mathrm{2}} }−{na}^{{n}} \:=\frac{{na}^{{n}} }{{a}−\mathrm{1}}−\frac{{a}^{{n}} −\mathrm{1}}{\left({a}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${n}=\mathrm{50} \\ $$$${S}=\mathrm{50}.\mathrm{2}^{\mathrm{50}} −\mathrm{2}^{\mathrm{50}} +\mathrm{1}=\mathrm{49}.\mathrm{2}^{\mathrm{50}} +\mathrm{1} \\ $$$$ \\ $$

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