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Step-by-Step Solution
Step 1: Write the General Term
Consider the binomial expansion of
(x + \frac{a}{x^2})^n .
The (r+1) -th term of this expansion can be written as:
T_{r+1} \;=\; \binom{n}{r} \;x^{\,n-r}\,\left(\frac{a}{x^2}\right)^{r}
\;=\;\binom{n}{r}\,a^r \,x^{\,n - 3r}.
Step 2: Identify the Coefficients of the 3rd, 4th, and 5th Terms
1. The 3rd term corresponds to r=2 , so
T_{3} = \binom{n}{2}\,a^2\,x^{\,n - 6}.
2. The 4th term corresponds to r=3 , so
T_{4} = \binom{n}{3}\,a^3\,x^{\,n - 9}.
3. The 5th term corresponds to r=4 , so
T_{5} = \binom{n}{4}\,a^4\,x^{\,n - 12}.
The coefficients (ignoring the x part) of these terms are:
Coefficient of T_{3} : \binom{n}{2}\,a^2
Coefficient of T_{4} : \binom{n}{3}\,a^3
Coefficient of T_{5} : \binom{n}{4}\,a^4
Step 3: Use the Given Ratio of Coefficients
The problem states that the coefficients of the 3rd, 4th, and 5th terms are in the ratio 12 : 8 : 3. We can use pairwise comparisons to find relations between n and a .
3rd to 4th Coefficient Ratio
Given
\frac{\text{Coefficient of }T_3}{\text{Coefficient of }T_4} = \frac{12}{8} = \frac{3}{2}.
So,
\frac{\binom{n}{2}\,a^2}{\binom{n}{3}\,a^3} = \frac{3}{2}.
Simplify the combination terms if necessary; however, a direct approach in many solutions is to interpret
\binom{n}{2} = \frac{n(n-1)}{2},
\binom{n}{3} = \frac{n(n-1)(n-2)}{6}.
But we keep track of n and a succinctly. Sometimes the simplification can be viewed as:
\frac{\binom{n}{2}}{\binom{n}{3}} = \frac{\frac{n(n-1)}{2}}{\frac{n(n-1)(n-2)}{6}} = \frac{6}{2(n-2)} = \frac{3}{n-2}.
Thus,
\frac{\binom{n}{2}\,a^2}{\binom{n}{3}\, a^3}
= \frac{3}{n-2} \cdot \frac{1}{a}
= \frac{3}{2}.
Hence,
\frac{3}{n-2} \cdot \frac{1}{a} = \frac{3}{2}
\;\Rightarrow\; \frac{1}{a(n - 2)} = \frac{1}{2}
\;\Rightarrow\; a(n - 2) = 2.
4th to 5th Coefficient Ratio
Similarly,
\frac{\text{Coefficient of }T_4}{\text{Coefficient of }T_5} = \frac{8}{3}.
So,
\frac{\binom{n}{3}\,a^3}{\binom{n}{4}\,a^4} = \frac{8}{3}.
Noting
\binom{n}{4} = \frac{n(n-1)(n-2)(n-3)}{24},
we get:
\frac{\binom{n}{3}}{\binom{n}{4}}
= \frac{\frac{n(n-1)(n-2)}{6}}{\frac{n(n-1)(n-2)(n-3)}{24}}
= \frac{24}{6(n-3)}
= \frac{4}{n-3}.
Thus,
\frac{\binom{n}{3}\,a^3}{\binom{n}{4}\, a^4}
= \frac{4}{n-3}\cdot \frac{1}{a}
= \frac{8}{3}.
Hence,
\frac{4}{n-3}\cdot \frac{1}{a} = \frac{8}{3}
\;\Rightarrow\; \frac{1}{a(n-3)} = \frac{2}{3}
\;\Rightarrow\; a(n - 3) = \frac{3}{2}.
Step 4: Solve for n and a
We have two equations:
a(n - 2) = 2 \quad (1)
a(n - 3) = \frac{3}{2} \quad (2)
Subtracting these or solving simultaneously, from (1)-(2) :
a(n - 2) - a(n - 3) = 2 - \frac{3}{2}.
a[(n - 2) - (n - 3)] = \frac{4 - 3}{2} = \frac{1}{2}.
a(1) = \frac{1}{2}
\;\Rightarrow\; a = \frac{1}{2}.
Then from a(n - 2) = 2 :
\frac{1}{2}(n - 2) = 2
\;\Rightarrow\; n - 2 = 4
\;\Rightarrow\; n = 6.
Step 5: Find the Term Independent of x
For the term to be independent of x , the exponent of x in the general term must be zero:
n - 3r = 0
\;\Rightarrow\; r = \frac{n}{3} = \frac{6}{3} = 2.
The term with r=2 is the third term (as we labeled earlier T_3 ). Now substitute n=6 and a=\frac{1}{2} :
T_{3}
= \binom{6}{2}
\left(\frac{1}{2}\right)^{2}
x^{\,6 - 6}
= \binom{6}{2}
\left(\frac{1}{4}\right)
x^0.
And
\binom{6}{2} = 15.
Thus,
T_{3} = 15 \times \frac{1}{4} = \frac{15}{4}.
Step 6: Final Answer
The term independent of x in the given expansion is
\frac{15}{4},
which can also be written as 3.75.
