Explanation:
C
We know binomial expansion of
\((1+a x)^n=1+{ }^n C_1(a x)+{ }^n C_2(a x)^2+\ldots \ldots . .\)
\(\text { Given, }{ }^n C_1(a x)=6 x\)
\(\text { and, }{ }^n C_2(a x)^2=16 x^2\)
\(\text { On comparing both sides, we get- }\)
\({ }^n C_1 a=6\)
\(\text { And, }{ }^n C_2 a^2=16\)
\(\text { Then, }\)
\(\frac{n !}{(n-1) ! 1 !} a=6 \Rightarrow n \cdot a=6\)
\(\frac{n !}{(n-2) ! 2 !} a^2=16 \Rightarrow \frac{n(n-1)}{2} a^2=16\)
\(\text { Now, } \quad \mathrm{n}^2 a^2=36\)
\(\quad a^2=\frac{36}{n^2}\)
\(\text { And, } \quad \frac{n(n-1)}{2} a^2=16\)
\(\frac{n(n-1)}{2 n^2}=\frac{16}{36}=\frac{4}{9}\)
\(\text { or } \frac{n-1}{n}=\frac{8}{9}\)
\(\text { or } \quad 1-\frac{1}{n}=\frac{8}{9}\)
\(\text { or } \quad \frac{1}{n}=1-\frac{8}{9}=\frac{1}{9}\)
\(\therefore \quad n=9\)
Now, \(\mathrm{n}^2 \mathrm{a}^2=36\)
And, \(\quad \frac{\mathrm{n}(\mathrm{n}-1)}{2} \mathrm{a}^2=16\)
Then, \(\quad 81 \mathrm{a}^2=36\)
\(\therefore \quad \mathrm{a}^2 =\frac{36}{81}\)
\(\mathrm{a} =\frac{6}{9}=\frac{2}{3}\)So, the value of a and \(n\) are \(2 / 3\) and 9 respectively.
