Solveeit Logo
Login

Question

Question: If then \[x = \dfrac{1}{5} + \dfrac{{1 \cdot 3}}{{5 \cdot 10}} + \dfrac{{1 \cdot 3 \cdot 5}}{{5 \cdo...

If then x=15+1⋅35⋅10+1⋅3⋅55⋅10⋅15+.....∞x = \dfrac{1}{5} + \dfrac{{1 \cdot 3}}{{5 \cdot 10}} + \dfrac{{1 \cdot 3 \cdot 5}}{{5 \cdot 10 \cdot 15}} + .....\infty then 3x2+6x=3{x^2} + 6x =
A. 1
B. 2
C. 3
D. 4

Explanation

Solution

The given series can be written in the form of Binomial Expansion i.e., (1+x)n=1+nx+n(n+1)2!x2+n(n+1)(n+2)3!x3{\left( {1 + x} \right)^n} = 1 + nx + \dfrac{{n\left( {n + 1} \right)}}{{2!}}{x^2} + \dfrac{{n\left( {n + 1} \right)\left( {n + 2} \right)}}{{3!}}{x^3} in which by taking common terms from the given series we need to find the value of 3x2+6x3{x^2} + 6x, hence from the equation we need to find nxnx which implies the value of x and then substitute the value of x and evaluate the equation.

Complete step by step solution:
The given series is
x=15+1⋅35⋅10+1⋅3⋅55⋅10⋅15+.....∞x = \dfrac{1}{5} + \dfrac{{1 \cdot 3}}{{5 \cdot 10}} + \dfrac{{1 \cdot 3 \cdot 5}}{{5 \cdot 10 \cdot 15}} + .....\infty
We know that, according to Binomial expansion:
(1+x)n=1+nx+n(n+1)2!x2+n(n+1)(n+2)3!x3{\left( {1 + x} \right)^n} = 1 + nx + \dfrac{{n\left( {n + 1} \right)}}{{2!}}{x^2} + \dfrac{{n\left( {n + 1} \right)\left( {n + 2} \right)}}{{3!}}{x^3} …………………… 1
Taking the common terms, the equation can be written as:
1+x=1+15+1⋅32!(15)2+1⋅3⋅53!(15)31 + x = 1 + \dfrac{1}{5} + \dfrac{{1 \cdot 3}}{{2!}}{\left( {\dfrac{1}{5}} \right)^2} + \dfrac{{1 \cdot 3 \cdot 5}}{{3!}}{\left( {\dfrac{1}{5}} \right)^3}
Here, as per the equation 1 the value of nxnx from given expansion is
nxnx= 15\dfrac{1}{5}
Which implies the value of x as:
x=15nx = \dfrac{1}{{5n}} ………………… 2

And as per the equation 1 the value of n(n+1)2!x2\dfrac{{n\left( {n + 1} \right)}}{{2!}}{x^2} from given series is
n(n+1)2!x2\dfrac{{n\left( {n + 1} \right)}}{{2!}}{x^2}= 32!(15)2\dfrac{3}{{2!}}{\left( {\dfrac{1}{5}} \right)^2}
Substitute the value of x from equation 2 as: x=15nx = \dfrac{1}{{5n}}
n(n+1)2!152n2=32!(15)2\dfrac{{n\left( {n + 1} \right)}}{{2!}}\dfrac{1}{{{5^2}{n^2}}} = \dfrac{3}{{2!}}{\left( {\dfrac{1}{5}} \right)^2}
⇒n(n+1)1n=3\Rightarrow n\left( {n + 1} \right)\dfrac{1}{n} = 3
Simplifying the terms to find the value of n i.e.,
(n+1)n=3\dfrac{{\left( {n + 1} \right)}}{n} = 3
⇒(n+1)=3n\Rightarrow \left( {n + 1} \right) = 3n

Therefore, the value of n is:
n=12n = \dfrac{1}{2}
Substitute the value of n in equation 2 i.e.,
x=15nx = \dfrac{1}{{5n}}
⇒x=15(12)=152\Rightarrow x = \dfrac{1}{{5\left( {\dfrac{1}{2}} \right)}} = \dfrac{1}{{\dfrac{5}{2}}}
Hence, the value of x is
x=25x = \dfrac{2}{5}
Therefore, we get
(1+x)n=(53)12{\left( {1 + x} \right)^n} = {\left( {\dfrac{5}{3}} \right)^{\dfrac{1}{2}}}
⇒x2+2x+1=53\Rightarrow{x^2} + 2x + 1 = \dfrac{5}{3}
Simplifying the terms, we get:
3x2+6x=23{x^2} + 6x = 2

Therefore, option B is the right answer.

Note: The key point to find the given series is, we must know Binomial expansions to find the given series, as the Binomial Theorem is the method of expanding an expression which has been raised to any finite power and by evaluating the expansion, we can find the value of equation asked. As the power increases the expansion becomes lengthy and tedious to calculate. A binomial expression that has been raised to a very large power can be easily calculated with the help of the Binomial Theorem.