Question

If ${{c}_{0}},{{c}_{1}},{{c}_{2}},.......$ are binomial coefficient in the expansion of ${{\left( 1+x \right)}^{n}}$ , then
$\dfrac{{{c}_{0}}}{3}-\dfrac{{{c}_{1}}}{4}+-+.......$ is equal to:
(a). $\dfrac{1}{n+1}-\dfrac{2}{n+2}+\dfrac{1}{n+3}$
(b). $\dfrac{1}{n+1}+\dfrac{2}{n+2}-\dfrac{3}{n+3}$
(c). $\dfrac{1}{n+2}-\dfrac{1}{n+1}+\dfrac{1}{n+3}$
(d). $\dfrac{2}{n+1}-\dfrac{1}{n+2}+\dfrac{2}{n+3}$
(e). $\dfrac{1}{n+2}-\dfrac{2}{n+1}+\dfrac{3}{n+3}$

Answer 1

Hint: We do this question in a very efficient and easy way, first we will put the value of n = 0, the only term we get is ${{c}_{0}}$ one, and again we will put n = 1 , then we will get ${{c}_{0}}\text{ }and\text{ }{{c}_{1}}$ , just by putting these two values we will try to find the value of the expression and then we will compare it the given options to point out the correct answer.

Complete step-by-step solution -
The binomial expansion of ${{\left( 1+x \right)}^{n}}$is:
${{c}_{0}}+{{c}_{1}}x+{{c}_{2}}{{x}^{2}}+.....$
Now if we put n = 0 we only get ${{c}_{0}}$,
Therefore ${{c}_{0}}$= 1, and other terms will be zero
$\dfrac{{{c}_{0}}}{3}-\dfrac{{{c}_{1}}}{4}+-+.......$ will be equal to $\dfrac{1}{3}$ .
And now we will again put n = 1,
Therefore we get ${{c}_{0}}$= 1 and ${{c}_{1}}=1$ ,
Now the value of the expression will be:
$\dfrac{{{c}_{0}}}{3}-\dfrac{{{c}_{1}}}{4}+-+.......$ = $\dfrac{1}{3}-\dfrac{1}{4}=\dfrac{1}{12}$ .
Now we have got the two values for n = 0 and n = 1, we will put this in all the given options and then check which gives the correct value for both the values of n.
Let’s check option (a),
For n = 0 we get,
$\dfrac{1}{n+1}-\dfrac{2}{n+2}+\dfrac{1}{n+3}$ = $\dfrac{1}{3}$
And for n = 1 we get,
$\dfrac{1}{n+1}-\dfrac{2}{n+2}+\dfrac{1}{n+3}$ = $\dfrac{1}{2}-\dfrac{2}{3}+\dfrac{1}{4}=\dfrac{1}{12}$
So, we get both the values correct.
Let’s check option (b),
For n = 0 we get,
$\dfrac{1}{n+1}+\dfrac{2}{n+2}-\dfrac{3}{n+3}$ = 1
Which is not what we expected,
So, option (b) is incorrect.
Let’s check option (c),
For n = 0 we get,
$\dfrac{1}{n+2}-\dfrac{1}{n+1}+\dfrac{1}{n+3}$
= $\dfrac{1}{2}-1+\dfrac{1}{3}=\dfrac{-1}{6}$
Which is not what we expected,
So, option (c) is incorrect.
Let’s check option (d),
For n = 0 we get,
$\dfrac{2}{n+1}-\dfrac{1}{n+2}+\dfrac{2}{n+3}$
$=2-\dfrac{1}{2}+\dfrac{2}{3}=\dfrac{13}{6}$
Which is not what we expected,
So, option (d) is incorrect.
Let’s check option (e),
For n = 0 we get,
$\dfrac{1}{n+2}-\dfrac{2}{n+1}+\dfrac{3}{n+3}$
$=\dfrac{1}{2}-2+1=\dfrac{-1}{2}$
Which is not what we expected,
So, option (e) is incorrect.
Hence the only option that we get the correct answer is (a).
So, option (a) is correct.

Note: We can also solve this question by writing the value of all the binomial coefficients and then putting it in the given expression and then we have to arrange it then cancel some terms to get the given expression as in the option.