Question

The sum of the series $1 + 3x + 6{x^2} + 10{x^3} + ..............\infty$ will be
$\left( 1 \right)$ $\dfrac{1}{{{{\left( {1 - x} \right)}^2}}}$
$\left( 2 \right)$ $\dfrac{1}{{\left( {1 - x} \right)}}$
$\left( 3 \right)$ $\dfrac{1}{{{{\left( {1 + x} \right)}^2}}}$
$\left( 4 \right)$ $\dfrac{1}{{{{\left( {1 - x} \right)}^3}}}$

Answer 1

We have to find the sum of the given infinite series . We solve this question using the concept of sum of infinite series of geometric progression ( G.P. ) . From the given values we can compute the first term $a$ and the common ratio $r$ . And using the formula of infinite series of G.P. we can find the expression of sum of the series .

Complete step-by-step solution:
Given :
$1 + 3x + 6{x^2} + 10{x^3} + ..............\infty$ is the given series
Let $S$ be is the sum of the given series
$S = 1 + 3x + 6{x^2} + 10{x^3} + ..............\infty - - - \left( 1 \right)$
Multiply both sides by $x$ , we can write the expression as :
$S \times x = x + 3{x^2} + 6{x^3} + 10{x^4} + ..............\infty - - - \left( 2 \right)$
subtracting equation $\left( 2 \right)$ from equation $\left( 1 \right)$ , we get the expression as :
$S - S \times x = 1 + 3x + 6{x^2} + 10{x^3} + ..............\infty - \left[ {x + 3{x^2} + 6{x^3} + 10{x^4} + ..............\infty } \right]$
$\left( {1 - x} \right) \times S = 1 + 2x + 3{x^2} + 4{x^3} + 5{x^4} + ..............\infty - - - \left( 3 \right)$
Multiply both sides by $x$ , we get the expression as :
$x \times \left( {1 - x} \right) \times S = x + 2{x^2} + 3{x^3} + 4{x^4} + 5{x^5} + ..............\infty - - - \left( 4 \right)$
Subtracting equation $\left( 4 \right)$ from equation $\left( 3 \right)$ , we get the expression as :
$\left( {1 - x} \right) \times S - \left[ {x \times \left( {1 - x} \right) \times S} \right] = 1 + 2x + 3{x^2} + 4{x^3} + 5{x^4} + ..............\infty - \left[ {x + 2{x^2} + 3{x^3} + 4{x^4} + 5{x^5} + ..............\infty } \right]$
${\left( {1 - x} \right)^2} \times S = 1 + x + {x^2} + {x^3} + {x^4} + {x^5} + ..............\infty$
Now , for the sum of infinite terms of G.P.
First term of the G.P. is given as :
$a = 1$
Common ratio of the series is given as :
$r = x$
We know , that the formula of sum of infinite series is given as :
${S_\infty } = \dfrac{a}{{1 - r}}$
Where $\left| r \right| < 1$ .
Putting the values in the formula of sum of infinite series and simplifying the expression , we get
$S = \dfrac{a}{{1 - x}}$
So , equating the value we can write the expression as :
${\left( {1 - x} \right)^2} \times S = \dfrac{a}{{1 - x}}$
$S = \dfrac{a}{{{{\left( {1 - x} \right)}^3}}}$
Thus , the sum of the series is $\dfrac{a}{{{{\left( {1 - x} \right)}^3}}}$ .
Hence , the correct option is $\left( 4 \right)$ .

Note: For the terms of a given series to be in G.P. the common ratio between the terms of the series should be the same for all the two consecutive terms of the series . The ratio of the second term to the first term of the given series should be the same as that of the ratio of the third term to the second term of the given series .
The sum of $n$ terms of a G.P. is given by the formula :
${S_n} = \dfrac{{a\left( {{r^n} - 1} \right)}}{{r - 1}}$
The expression for ${n^{th}}$ term of G.P. is given by the formula :
${a_n} = a \times {r^{n - 1}}$