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Question: If the third term in the binomial expansion of \[{{\left( 1+x \right)}^{m}}\text{ is }-\dfrac{1}{8}{...

If the third term in the binomial expansion of (1+x)m is −18x2{{\left( 1+x \right)}^{m}}\text{ is }-\dfrac{1}{8}{{x}^{2}}, then the rational value of m is:
(a) 2
(b) 12\dfrac{1}{2}
(c) 3
(d) 4

Explanation

Solution

Hint: First find the binomial expansion of (1+x)m{{\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:
(a+b)n=an+ nC1an−1b+.....+ nCran−rbr+.....+ nCn−1abn−1+bn{{\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
(a+b)1=a1+b1{{\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
(a+b)2=a2+ 2C1a2−1b+b2{{\left( a+b \right)}^{2}}={{a}^{2}}+{{\text{ }}^{2}}{{C}_{1}}{{a}^{2-1}}b+{{b}^{2}}
By simplifying, we get,
(a+b)2=a2+2ab+b2{{\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,
(a+b)k=ak+ kC1ak−1b1+ kC2ak−2b2+.....+ kCkabk−1+bk{{\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:
(a+b)k+1=(a+b)(a+b)k{{\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:
(a+b)(ak+ kC1ak−1b+ kC2ak−2b2+......+bk)\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,
(a+b).c=a.c+b.c\left( a+b \right).c=a.c+b.c
⇒ak+1+(1+ kC1)akb+(kC1+ kC2)ak−1b2+.....+bk+1\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:
nCr−1+ nCr= n+1Cr^{n}{{C}_{r-1}}+{{\text{ }}^{n}}{{C}_{r}}={{\text{ }}^{n+1}}{{C}_{r}}
By using Pascal’s identity, we get,
(a+b)k+1=ak+1+ kC1akb+.....+ k+1Crak+1−rbr+.....+bk+1{{\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
(a+b)n=∑k=0nnCkakbn−k{{\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:
(1+x)m=1+mx+m(m−1)2x2+.....{{\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
m(m−1)2x2\dfrac{m\left( m-1 \right)}{2}{{x}^{2}}
The third term given in the question can be written as:
−18x2\dfrac{-1}{8}{{x}^{2}}
By equating both the terms, we get the equation as:
m(m−1)2x2=−18x2\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:
m(m−1)2=−18\dfrac{m\left( m-1 \right)}{2}=\dfrac{-1}{8}
By multiplying with 2 on both the sides, we get it as:
m(m−1)=−14m\left( m-1 \right)=\dfrac{-1}{4}
By multiplying with 4 on both the sides, we get it as:
4m(m−1)=−14m\left( m-1 \right)=-1
By applying distributive law on the left-hand side, we get,
4m2−4m=−14{{m}^{2}}-4m=-1
By adding 1 on both the sides of the equation, we get it as:
4m2−4m+1=04{{m}^{2}}-4m+1=0
Comparing this to the equation ax2+bx+ca{{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:
4m2−2m−2m+1=04{{m}^{2}}-2m-2m+1=0
By taking 2m common from the first two, – 1 common from the last two, we get
2m(2m−1)−1(2m−1)=02m\left( 2m-1 \right)-1\left( 2m-1 \right)=0
We can write the equation as product form, given by:
(2m−1)(2m−1)=0\left( 2m-1 \right)\left( 2m-1 \right)=0
By equating the expression to 0, we get it as:
2m−1=02m-1=0
By simplifying we say the value of m to be as:
m=12m=\dfrac{1}{2}
By simplifying the given condition, we need the value of m to be 12\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 a2−2ab+b2=(a−b)2{{a}^{2}}-2ab+{{b}^{2}}={{\left( a-b \right)}^{2}} formula instead of factorization. Anyways you will get the same result.