Home
Class 12
MATHS
Find the sum of the series 1+2(1-x)+3(1-...

Find the sum of the series `1+2(1-x)+3(1-x)(1-2x)++n(1-x)(1-2x)(1-3x)[1-(n-1)x]` .

Text Solution

Verified by Experts

The correct Answer is:
`1/x[1-(1-x)(1-2x)...(1-nx)`
`T_(r)=r(1-x)(1-2x)(1-3x)…(1-(r-1)x)`
`=(1-x)(1-2x)(1-3x)…(1-(r-1)x)(xr-1+1)/x`
`=1/x{(1-x)(1-2x)…(1-(r-1)x)-(1-x)(1-2x)..(1-(r-1)x)(1-rx)}`
`=1/x[V(r-1)-V(r )]`, where
`V(r )=(1-x)(1-2x)…(1-rx)`
`therefore` Required sum=`sum_(r=1)^(n)T_(r)=1/xsum_(r=1)^(n)V(r-1)-V( r)`
`=1/x(V(0)-V(n))`
`1/x[1-(1-x)(1-2x)...(1-nx)]`
Promotional Banner

Topper's Solved these Questions

  • PROGRESSION AND SERIES

    CENGAGE|Exercise Exercise (Single)|93 Videos

  • PROGRESSION AND SERIES

    CENGAGE|Exercise EXERCIESE ( MULTIPLE CORRECT ANSWER TYPE )|1 Videos

  • PROGRESSION AND SERIES

    CENGAGE|Exercise Exercise 5.8|10 Videos

  • PROBABILITY II

    CENGAGE|Exercise NUMARICAL VALUE TYPE|2 Videos

  • PROPERTIES AND SOLUTIONS OF TRIANGLE

    CENGAGE|Exercise JEE Advanced Previous Year|11 Videos

Similar Questions

Explore conceptually related problems

Find the sum of the series e^(x-(1)/(2)(x -1)^(2) + (1)/(3) (x -1)^(3) - (1)/(4) (x -1)^(4) + ...

For x>0 the sum of the series (1)/(1+x)-(1-x)/((1+x)^(2))+((1-x)^(2))/((1+x)^(3))-...oo is equal to

Sum to n terms of the series (1)/((1+x)(1+2x))+(1)/((1+2x)(1+3x))+(1)/((1+3x)(1+4x))+ ………

Sum to n terms of the series (1)/((1+x)(1+2x))+(1)/((1+2x)(1+3x))+(1)/((1+3x)(1+4x))+ is (i) (1+x)(1+nx)((1+x)(1+nx))/((1+x)(1+(n-1)x))( iv )((1+x)(1+(n+1)x))/((1+x)(1+(n+1)x))

Find the sum of the series 1+3x+5x^(2)+7x^(2)+... to n terms.

Find the sum of the series nC_(0)x^(3n)+(nC_(1))/(2)x^(3(n-1))(3-x^(3))+(nC_(2))/(3)x^(3(n-2))(3-x^(3))^(2)+......+(nC_(n))/(n+1)(3-x^(3))^(n)

Sum to n terms of the series 1+(1+x)+(1+x+x^(2))+(1+x+x^(2)+x^(3))+

Find the coefficient of x^(n) in the expansion of (1)/((1-x)(1-2x)(1-3x))

Find the sum to n terms of the series: 1+2x+3x^(2)+4x^(3)+......

Find the sum of the series 1+2x+3x^(2)+(n-1)x^(n-2) using differentiation.