JEE Main & Advanced Mathematics Sequence & Series Question Bank Arithmetico - Geometric Series, Method of difference

  • question_answer
    The sum of the series \[1+3x+6{{x}^{2}}+10{{x}^{3}}+........\infty \]  will be

    A) \[\frac{1}{{{(1-x)}^{2}}}\]

    B) \[\frac{1}{1-x}\]

    C) \[\frac{1}{{{(1+x)}^{2}}}\]

    D) \[\frac{1}{{{(1-x)}^{3}}}\]

    Correct Answer: D

    Solution :

    Let \[S=1+3x+6{{x}^{2}}+10{{x}^{3}}+.....\infty \] \[\Rightarrow \]\[x.S=x+3{{x}^{2}}+6{{x}^{3}}+.......\infty \] Subtracting \[S(1-x)=1+2x+3{{x}^{2}}+4{{x}^{3}}+.......\infty \] \[\Rightarrow \]\[x(1-x)S=x+2{{x}^{2}}+3{{x}^{3}}+.......\infty \] Again subtracting, \[\Rightarrow \]\[S[(1-x)-x(1-x)]=1+x+{{x}^{2}}+{{x}^{3}}+........\infty \] \[\Rightarrow \]\[S[(1-x)(1-x)]=\frac{1}{1-x}\Rightarrow S=\frac{1}{{{(1-x)}^{3}}}\]


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