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InterviewSolution
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Find the sum of `n`terms of the series `1+4/5+7/(5^2)+10/5^3+dot` |
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Answer» The given series can be written as `1+4((1)/(5))+7((1)/(5))^(2)+10((1)/(5))^(3)+"....."` The series is an Arithmetico-Geometric series, since each term is formed by multiplng corresponding terms of the series `1,4,7,"….."` which are in AP and `1,(1)/(5),(1)/(5)^(2),"...."` which are in GP. ` therefore T_(n)=[" nth term of "1,4,7,"..."][" n th term of " 1,(1)/(5),)((1)/(5))^(2)+"....."]` `=[1+(n-1)3]xx1*((1)/(5))^(n-1)=(3n-2)((1)/(5))^(n-1)"...(i)"` ` therefore T_(n-1)=(3n-5)((1)/(5))^(n-2)` (i) Let sum of n terms of the series is denoted by `S_(n)` Then, ` S_(n)=1+4((1)/(5))+7((1)/(5))^(2)+"...."+(3n-5)((1)/(5))^(n-2)+(3n-2)((1)/(5))^(n-1)" ......(i)"` Multiplying both the sides of Eq. (i) by `(1)/(5)`, we get ` (1)/(5)S_(n)=(1)/(5)+4((1)/(5))^(2)+7((1)/(5))^(3)+"...."+(3n-5)((1)/(5))^(n-1)+(3n-2)((1)/(5))^(n)"......(ii)"` Subtracting Eq. (ii) from Eq.(i), we get ` (1-(1)/(5))S_(n)=1+3[(1)/(5)+((1)/(5))^(2)+((1)/(5))^(3)+"...."+((1)/(5))^(n-1)]-(3n-2)((1)/(5))^(n)` or ` (4)/(5)S_(n)=1+3[((1)/(5))+((1)/(5))^(2)+((1)/(5))^(3)+"...."+(n-1)" terms "]-(3n-2)((1)/(5))^(n)` ` =1+3[[(1)/(5)[1-((1)/(5))^(n-1)])/(1-(1)/(5))]-(3n-2)((1)/(5))^(n)` ` =1+(3)/(4)[1-((1)/(5))^(n-1)]-(3n-2)((1)/(5))^(n)` ` therefore S_(n) =(5)/(4)+(15)/(16)[1-((1)/(5))^(n-1)]-((3n-2))/(4)((1)/(5))^(n-1)` `=(35)/(16)-((12n+7))/(16)((1)/(5))^(n-1)` |
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