Mathematics Test - 48

The 7 th term of the given binomial expansion will be, \(={ }_{6}^{13} \mathrm{C}(4 x)^{7}\left(-\frac{1}{2 x^{\frac{1}{2}}}\right)^{6}\)
\(=\frac{13 ! \times 2^{14} \times x^{7}}{7 ! \cdot 6 ! \times 2^{6} \cdot x^{3}}\)
\(=\frac{13 ! \times 2^{14}}{7 ! \times 6 ! \times 2^{6}} x^{4}\)
\(=439296 x^{4}\)
Hence, the correct option is (B)

The coefficient of \(x^{m}\) will be summation of \({ }^{m} C_{m}+{ }^{m+1} C_{m}+{ }^{m+2} C_{m} \ldots{ }^{n} C_{m}\)
Upon simplifying we get \(C_{m+1}\)
Hence, the correct option is (A)

Consider \((1-x)^{20}\)
\(=1-{ }^{20} C_{1} x^{1}+{ }^{20} C_{2} x^{2}-{ }^{20} C_{3} x^{3} \ldots+{ }^{20} C_{20} x^{20}\)
By differentiating with respect to x, we get \(-n(1-x)^{n-1}=-20 C_{1}+2^{20} C_{2} x^{1}-3^{20} C_{3} x^{2}+\ldots .20^{20} C_{20} x^{19}\)
Substituting \(\mathrm{x}=\mathrm{I},\) we get \(0=-{ }^{20} C_{1}+2{ }^{20} C_{2}-3{ }^{20} C_{3}+\ldots .20^{20} C_{20}\)
Hence answer is \(0 .\)
Hence, the correct option is (B)

Given, The sum of all binomial coefficients is 256 So, \(2^{n}=256=2^{8}\) and \(n=8\)
The term with the greatest binomial coefficient is the middle term that is \(T_{5}\)
\(T_{5}=\frac{4}{8} C\left(3 \frac{-x}{4}\right)^{4}\left(3 \frac{5 x}{4}\right)^{4}\)
and third term, \(T_{3}={ }_{8}^{2} C\left(3 \frac{-x}{4}\right)^{6}\left(3 \frac{5 x}{4}\right)^{2}\)
Given,

\((1+x)^{n}(1+y)^{n}(1+z)^{n}\)
\(=\left(\sum_{r=0}^{n}{ }^{n} C_{r} x^{r}\right)\left(\sum_{s=0}^{n}{ }^{n} C_{s} y^{s}\right)\left(\sum_{t=0}^{n}{ }^{n} C_{t} z^{t}\right)\)
\(=\sum_{0 \leq r, s, t \leq n}\left({ }^{n} C_{r}\right)\left({ }^{n} C_{s}\right)\left({ }^{n} C_{t}\right) x^{r} y^{s} z^{t}\)
For sum of the coefficients of degree \(m,\) we must have \(r+s+t=m\) Where \(r, s, t\) are integers with \(r, s, t \geq 0\) Sum of such coefficients \(=\sum_{r, s, t \geq 0 r+s+t=m}\left({ }^{n} C_{r}\right)\left({ }^{n} C_{s}\right)\left({ }^{n} C_{t}\right)\)
=the number of ways of choosing a total number of \(m\) balls out of \(n\) black, \(n\) white and \(n\) green balls \(={ }^{3 n} C_{m}\)
Hence, the correct option is (C)

\(\left(1+x+2 x^{2}\right)^{20}=a_{0}+a_{1} x+a_{2} x^{2}+\ldots \ldots \ldots+a_{40} x^{40}\)
Put \(x=1\) in ( 1\()\) \(4^{20}=a_{0}+a_{1}+a_{2}+\ldots \ldots \ldots+a_{40}\)
\(\Rightarrow 2^{40}=a_{0}+a_{1}+a_{2}+\ldots \ldots \ldots+a_{40} \ldots .(2)\)
Put \(x=-1\) in ( 1\()\) \(\Rightarrow 2^{20}=a_{0}-a_{1}+a_{2}-a_{3} \ldots \ldots \ldots a_{39}+a_{40} \ldots .(3)\)
Adding ( 2 ) and ( 3 ), we get \(2^{40}+2^{20}=2\left(a_{0}+a_{2}+a_{4}+\ldots \ldots+a_{38}+a_{40}\right)\)
\(2^{20}\left(2^{20}+1\right)=2\left(a_{0}+a_{2}+a_{4}+\ldots \ldots+a_{38}+a_{40}\right)\)
\(\Rightarrow 2^{19}\left(2^{20}+1\right)=a_{0}+a_{2}+a_{4}+\ldots \ldots+a_{38}+a_{40}\)
\(a_{0}+a_{2}+a_{4}+\ldots \ldots+a_{38}=2^{19}\left(2^{20}+1\right)-a_{40} \ldots . .(4)\)
since, \(a_{40}\) is the coefficient of \(x^{40}\) in the expansion of \(\left(1+x+2 x^{2}\right)^{20}\) \(\Rightarrow a_{40}=2^{20}\)
Put this value in (4), we get \(a_{0}+a_{2}+a_{4}+\ldots \ldots+a_{38}=2^{19}\left(2^{20}+1\right)-2^{20}\)
\(\Rightarrow a_{0}+a_{2}+a_{4}+\ldots \ldots+a_{38}=2^{19}\left(2^{20}-1\right)\)
Hence, the correct option is (A)

The given summation can be re-written as \(\sum_{r=0}^{n-1} \frac{{ }^{n} C_{r}}{{ }_{n+1} C_{r+1}} \ldots\) (Using the
properties of binomial coefficient) \(\rightarrow \sum_{r=0}^{n-1} \frac{r+1}{n+1}\)
\(=\frac{1}{n+1}+\frac{2}{n+1}+\frac{3}{n+1}+\ldots \frac{n}{n+1}\)
\(=\frac{n(n+1)}{2(n+1)}\)
\(=\frac{n}{2}\)
Hence, the correct option is (A)

Given, A is circulant matrix of elements \(a, b, c\) and \(a b c=1\) \(\mathrm{So}, A=\left[\begin{array}{lll}a & b & c \\ c & a & b \\ b & c & a\end{array}\right]\)
So \(\operatorname{det} A=a\left(a^{2}-b c\right)-b\left(a c-b^{2}\right)+c\left(c^{2}-a b\right)\)
\(=a^{3}+b^{3}+c^{3}-3 a b c\)
det \(A=a^{2}+b^{3}+c^{3}-3\)
and \(A^{T} A=I\)
\(\left|A^{T} A\right|=|I|=1\)
\(|A|\left|A^{T}\right|=1\)
\(|A|^{2}=1\)
\(|A|=\pm 1\)
So, \(\operatorname{det} A=\pm 1\)
After substituing the value in the
\(\operatorname{det} A=a^{3}+b 63+c^{3}-3 a b c\)
we get:
\(a^{3}+b^{3}+c^{3}=4\) or
\(a^{3}+b^{3}+C^{3}=-2\)
Hence, the correct option is (D)