If the third term in the binomial expansion of \[{{\left( 1+x \right)}^{m}}\text{ is }-\dfrac{1}{8}{{x}^{2}}\], then the rational value of m is:
(a) 2
(b) \[\dfrac{1}{2}\]
(c) 3
(d) 4

Hint: First find the binomial expansion of \[{{\left( 1+x \right)}^{m}}\]. Now compare the third term with the given value. Cancel the common terms to get the value of m. The equation will be a single variable equation of m. The value of m is the required result.

Complete step-by-step answer:
Proof of binomial theorem by Mathematical Induction: The expression, we aim to prove is:
\[{{\left( a+b \right)}^{n}}={{a}^{n}}+{{\text{ }}^{n}}{{C}_{1}}{{a}^{n-1}}b+.....+{{\text{ }}^{n}}{{C}_{r}}{{a}^{n-r}}{{b}^{r}}+.....+{{\text{ }}^{n}}{{C}_{n-1}}a{{b}^{n-1}}+{{b}^{n}}\]
We will first prove that the relationship is true for n = 1, n = 2.
Case 1: n = 1
\[{{\left( a+b \right)}^{1}}={{a}^{1}}+{{b}^{1}}\]
This is true as the left-hand side is equal to the right-hand side.
Case 2: n = 2
\[{{\left( a+b \right)}^{2}}={{a}^{2}}+{{\text{ }}^{2}}{{C}_{1}}{{a}^{2-1}}b+{{b}^{2}}\]
By simplifying, we get,
\[{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}\]
This is true as it is a general algebraic identity.
Now, we should take the case where n = k
By substituting this, we get,
\[{{\left( a+b \right)}^{k}}={{a}^{k}}+{{\text{ }}^{k}}{{C}_{1}}{{a}^{k-1}}{{b}^{1}}+{{\text{ }}^{k}}{{C}_{2}}{{a}^{k-2}}{{b}^{2}}+.....+{{\text{ }}^{k}}{{C}_{k}}a{{b}^{k-1}}+{{b}^{k}}\]
Now, consider the expansion:
\[{{\left( a+b \right)}^{k+1}}=\left( a+b \right){{\left( a+b \right)}^{k}}\]
By substituting the case of n = k into this equation, we get:
\[\left( a+b \right)\left( {{a}^{k}}+{{\text{ }}^{k}}{{C}_{1}}{{a}^{k-1}}b+{{\text{ }}^{k}}C_{2}{{a}^{k-2}}{{b}^{2}}+......+{{b}^{k}} \right)\]
By using distributive law, we get,
\[\left( a+b \right).c=a.c+b.c\]
\[\Rightarrow {{a}^{k+1}}+\left( 1+{{\text{ }}^{k}}{{C}_{1}} \right){{a}^{k}}b+\left( ^{k}{{C}_{1}}+{{\text{ }}^{k}}{{C}_{2}} \right){{a}^{k-1}}{{b}^{2}}+.....+{{b}^{k+1}}\]
We know Pascal’s identity on combinations is:
\[^{n}{{C}_{r-1}}+{{\text{ }}^{n}}{{C}_{r}}={{\text{ }}^{n+1}}{{C}_{r}}\]
By using Pascal’s identity, we get,
\[{{\left( a+b \right)}^{k+1}}={{a}^{k+1}}+{{\text{ }}^{k}}{{C}_{1}}{{a}^{k}}b+.....+{{\text{ }}^{k+1}}{{C}_{r}}{{a}^{k+1-r}}{{b}^{r}}+.....+{{b}^{k+1}}\]
So, by assuming n = k is correct, we have proved n = k + 1 correct as well.
So, the binomial theorem is proved by Mathematical Induction. According to the binomial theorem, we have
\[{{\left( a+b \right)}^{n}}=\sum\limits_{k=0}^{n}{^{n}{{C}_{k}}}{{a}^{k}}{{b}^{n-k}}\]
From the above binomial proof, we get the expansion as:
\[{{\left( 1+x \right)}^{m}}=1+mx+\dfrac{m\left( m-1 \right)}{2}{{x}^{2}}+.....\]
The third term of the above expression can be given as
\[\dfrac{m\left( m-1 \right)}{2}{{x}^{2}}\]
The third term given in the question can be written as:
\[\dfrac{-1}{8}{{x}^{2}}\]
By equating both the terms, we get the equation as:
\[\dfrac{m\left( m-1 \right)}{2}{{x}^{2}}=\dfrac{-1}{8}{{x}^{2}}\]
By canceling the common terms on both the sides of the equation, we get:
\[\dfrac{m\left( m-1 \right)}{2}=\dfrac{-1}{8}\]
By multiplying with 2 on both the sides, we get it as:
\[m\left( m-1 \right)=\dfrac{-1}{4}\]
By multiplying with 4 on both the sides, we get it as:
\[4m\left( m-1 \right)=-1\]
By applying distributive law on the left-hand side, we get,
\[4{{m}^{2}}-4m=-1\]
By adding 1 on both the sides of the equation, we get it as:
\[4{{m}^{2}}-4m+1=0\]
Comparing this to the equation \[a{{x}^{2}}+bx+c\], we can say a = 4, b = – 4, c = 1.
The product of two numbers ac value is +4. So, the two numbers whose product is 4, the sum is – 4 are – 2, – 2. By using these, we can write our equation as:
\[4{{m}^{2}}-2m-2m+1=0\]
By taking 2m common from the first two, – 1 common from the last two, we get
\[2m\left( 2m-1 \right)-1\left( 2m-1 \right)=0\]
We can write the equation as product form, given by:
\[\left( 2m-1 \right)\left( 2m-1 \right)=0\]
By equating the expression to 0, we get it as:
\[2m-1=0\]
By simplifying we say the value of m to be as:
\[m=\dfrac{1}{2}\]
By simplifying the given condition, we need the value of m to be \[\dfrac{1}{2}\].
Therefore, option (b) is the right answer.

Note: While expanding in binomial expansion, be careful with negative signs. After substituting in the third term, while multiplying take care of – 1 also. After getting the quadratic, you can use \[{{a}^{2}}-2ab+{{b}^{2}}={{\left( a-b \right)}^{2}}\] formula instead of factorization. Anyways you will get the same result.