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Find the sum of the series 1+2^(2)x+3^(2...

Find the sum of the series `1+2^(2)x+3^(2)x^(2)+4^(2)x^(3)+"...."" upto "infty|x|lt1`.

Text Solution

Verified by Experts

Here, the numbers `1^(2),2^(2),3^(2),4^(2),"...."i.e.1,4,9,16"...."` are not in AP but ` 1,4-1=3,9-4=5,16-9=7,"...."` are in AP.
Let `S_(infty)=1+2^(2)x+3^(2)x^(2)+4^(2)x^(3)+"...."" upto "infty`
`=1+4x+9x^(3)+16x^(3)+"...."" upto "infty "......(i)"`
Multiplying both sides of Eq. (i) by x, we get
`S_(infty)=x+4x^(2)+9x^(3)+16x^(4)+"...."" upto "infty "......(ii)"`
Subtracting Eq.(ii) from Eq.(i), we get
`(1-x)S_(infty)=1+3x+5x^(2)+7x^(3)+"...."" upto "infty "......(iii)"`
Again, multiplying both sides of Eq. (iii) by x, we get
`x(1-x)S_(infty)=x+3x^(2)+5x^(3)+7x^(4)+"...."" upto "infty "......(iv)"`
Subtracting Eq. (iv) from Eq. (iii),we get `(1-x)(1-x)S_(infty)=1+2x+2x^(2)+2x^(3)+"...."" upto "infty`
`=1+2(x+x^(2)+x^(3)+"...."" upto "infty)`
`=1+2((x)/(1-x))=((1+x)/(1-x))`
` therefore S_(infty)=((1+x))/((1-x)^(3))`
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Knowledge Check

  • The sum of the series 1 + 2x + 3x^(2) + 4x^(3) +"…." to oo where x lies between 0 and 1 i.e., 0 lt x lt 1 is :

    A
    `( 1)/( 1+x)`
    B
    `( 1)/( 1-x)`
    C
    `( 1)/( 1-2x)`
    D
    `( 1)/((1-x)^(2))`

  • Sum of n terms of the series 1+2 x+3 x^(2)+4 x^(3)+..,(x ne 1) is

    A
    1)`(1-x^(n))/((1-x)^(2))-(n x^(n))/(1-x)`
    B
    2)`(1+x^(n))/((1+x)^(2))-(n x^(n))/(1-x)`
    C
    3)`((1-x)^(n))/((1+x)^(2))-(n x^(n))/(1+x)`
    D
    4)`(1+x^(n))/((1-x)^(n))-(n x^(n))/(1-x)`

  • The sum of first 10 terms of the series : (x+(1)/( x))^(2) + ( x^(2) + ( 1)/( x^(2)))^(2) + ( x^(3)+ ( 1)/( x^(3)))^(2) + "......" is :

    A
    `( (x^(20)-1)/(x^(2) - 1))((x^(22) + 1)/( x^(20))) + 20`
    B
    `( (x^(18)-1)/(x^(2) - 1))((x^(11) + 1)/( x^(9))) + 20`
    C
    `( (x^(18)-1)/(x^(2) - 1))((x^(11) - 1)/( x^(9))) + 20`
    D
    None of these

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