
Find the value of the sum of expression $1+3x+6{{x}^{2}}+10{{x}^{3}}+.........$ to infinity
Answer
611.7k+ views
Hint: As there are sum up to infinite terms. In our knowledge we can find a sum of infinite terms if the series is geometric progression. So, by multiplying $x$ or dividing it, converts the given series into the form of geometric progression. So, if you multiply the series by $x$ and subtract this from the original series you get a series with decreased coefficients. Now repeat the same process once again then the series will get converted into a geometrical progression series. By applying the formula of sum of infinite geometric progression with first term “a”, common ratio “r”, sum “s” is: $s=\dfrac{a}{1-r}$.
Complete step-by-step answer:
Given expression, in the question for which we have to find the value is given by as follows:
$1+3x+6{{x}^{2}}+10{{x}^{3}}+.........\infty $
Let this expression be represented by S, we can write it as:
$s=1+3x+6{{x}^{2}}+.........+\infty ............(i)$
By multiplying with $''x''$ on both sides, we convert this into:
$sx=0+x+3{{x}^{2}}+.............\infty ..........(ii)$
By subtracting equation (ii) from equation (i), we convert this into:
$s\left( 1-x \right)=1+2x+3{{x}^{2}}+4{{x}^{3}}+.........\infty ..........(iii)$
By multiplying this equation with $x$ on both sides, we get:
$sx\left( 1-x \right)=0+x+2{{x}^{2}}+3{{x}^{3}}+.........+\infty ............(iv)$
By subtracting the equation (iv) from equation (iii) we convert this:
$s\left( \left( 1-x \right)-x\left( 1-x \right) \right)=1+x+{{x}^{2}}+{{x}^{3}}+........\infty $
By basic knowledge of algebra, we can write above as:
$1-x-\left( 1-x \right)x=1-2x+{{x}^{2}}={{\left( 1-x \right)}^{2}}$
By substituting this in above equation, we convert it into:
$s{{\left( 1-x \right)}^{2}}=1+x+{{x}^{2}}+.......\infty $
We can see right hand side is geometric progression expression,
By basic knowledge of progression, sum of infinite geometric progression is: $s=\dfrac{a}{1-r}$
Here, a = 1, $r = x$
So, above equation can be written as:
$s{{\left( 1-x \right)}^{2}}=\dfrac{1}{\left( 1-x \right)}$
By simplifying above equation, we can write it as:
$s=\dfrac{1}{{{\left( 1-x \right)}^{3}}}$
Hence value of required expression is $\dfrac{1}{{{\left( 1-x \right)}^{3}}}$
Note: The observation of multiplying with $x$ and subtracting is crucial as it takes us to result. Don’t confuse the idea.
Complete step-by-step answer:
Given expression, in the question for which we have to find the value is given by as follows:
$1+3x+6{{x}^{2}}+10{{x}^{3}}+.........\infty $
Let this expression be represented by S, we can write it as:
$s=1+3x+6{{x}^{2}}+.........+\infty ............(i)$
By multiplying with $''x''$ on both sides, we convert this into:
$sx=0+x+3{{x}^{2}}+.............\infty ..........(ii)$
By subtracting equation (ii) from equation (i), we convert this into:
$s\left( 1-x \right)=1+2x+3{{x}^{2}}+4{{x}^{3}}+.........\infty ..........(iii)$
By multiplying this equation with $x$ on both sides, we get:
$sx\left( 1-x \right)=0+x+2{{x}^{2}}+3{{x}^{3}}+.........+\infty ............(iv)$
By subtracting the equation (iv) from equation (iii) we convert this:
$s\left( \left( 1-x \right)-x\left( 1-x \right) \right)=1+x+{{x}^{2}}+{{x}^{3}}+........\infty $
By basic knowledge of algebra, we can write above as:
$1-x-\left( 1-x \right)x=1-2x+{{x}^{2}}={{\left( 1-x \right)}^{2}}$
By substituting this in above equation, we convert it into:
$s{{\left( 1-x \right)}^{2}}=1+x+{{x}^{2}}+.......\infty $
We can see right hand side is geometric progression expression,
By basic knowledge of progression, sum of infinite geometric progression is: $s=\dfrac{a}{1-r}$
Here, a = 1, $r = x$
So, above equation can be written as:
$s{{\left( 1-x \right)}^{2}}=\dfrac{1}{\left( 1-x \right)}$
By simplifying above equation, we can write it as:
$s=\dfrac{1}{{{\left( 1-x \right)}^{3}}}$
Hence value of required expression is $\dfrac{1}{{{\left( 1-x \right)}^{3}}}$
Note: The observation of multiplying with $x$ and subtracting is crucial as it takes us to result. Don’t confuse the idea.
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