Binomial Theorem Test - 77

Given the exprission is \(\left(2 \mathrm{x}+\frac{1}{\mathrm{x}}\right)^{8}\).

Here \(n=8(\mathrm{n}\) is even number)

\(\therefore\) Middle term \(=\left(\frac{\mathrm{n}}{2}+1\right)=\left(\frac{8}{2}+1\right)=5\) th term

General term in the expansion of \((x+y)^{n}\) is given by

\(\mathrm{T}_{(\mathrm{r}+1)}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} \times \mathrm{x}^{\mathrm{n}-\mathrm{r}} \times \mathrm{y}^{\mathrm{r}}\)

\(\Rightarrow T_{5}=T_{(4+1)}={ }^{8} \mathrm{C}_{4} \times(2 \mathrm{x})^{(8-4) \times}\left(\frac{1}{\mathrm{x}}\right)^{4}\)

\(\Rightarrow\mathrm{~T}_{5}={ }^{8} \mathrm{C}_{4} \times 2^{4}\)

We know that \((1+x)^{n}=C_{0}+C_{1} x+C_{2} x^{2}+\ldots+C_{n} x^{n}\) Differentiating both sides of the equation, we get:

\(n(1+x)^{n-1}=C_{1}+2 C_{2} x+3 C_{3} x^{2}+\ldots+n C_{n} x^{n-1}\)

Putting \(x=1\) and \(n=10\), we get:

\(\Rightarrow 10(1+1)^{10-1}=C_{1}+2 C_{2}+3 C_{3}+\ldots+10 C_{10}\)

\(\Rightarrow 10\left(2^{9}\right)=C_{1}+2 C_{2}+3 C_{3}+\ldots+10 C_{10}\)

\(\Rightarrow 10\left(2^{9}\right)=0 C_{0}+1 C_{1}+2 C_{2}+3 C_{3}+\ldots+10 C_{10}\)

\(\Rightarrow 10\left(2^{9}\right)=\sum_{\mathrm{r}=0}^{\mathrm{r}=10}\left(\mathrm{r} \times{ }^{10} \mathrm{C}_{\mathrm{r}}\right)\)

Given:

The \(6^{\text {th }}\) term in the expansion of \(\left[\frac{1}{x^{\frac{8 }{ 3}}}+x^{2} \log _{10} \mathrm{x}\right]^{8}\) is 5600......(1)

Using binomial theorem, the 6th term of \((a+b)^{8}\) is \(56 a^{3} b^{5}\)......(2)

Let \(a=x^{\frac{8 }{ 3}},~b=x^{2} \log _{10} \mathrm{x}\)

Using (1) and (2). we get 

\(56 \cdot \frac{1}{\left(\mathrm{x}^{\frac{8 }{ 3}}\right)^{3}} \cdot \mathrm{x}^{10} \cdot(\log \mathrm{x})^{5}=5600\)

\(\Rightarrow 56 \cdot \frac{1}{x^{8}} \cdot x^{10} \cdot(\log x)^{5}=5600\)

\(\Rightarrow x^{2}(\log x)^{5}=100\), \(x^{2}(\log x)^{5}=10^{2}\)

\(\Rightarrow x^{\frac{2}{5} \cdot 5}(\log x)^{5}=10^{2}\)

\(\Rightarrow \left(x^{\frac{2}{5}} \log x\right)^{5}=10^{2}\)

\(\Rightarrow \mathrm{x}^{\frac{2}{5}} \log \mathrm{x}=10^{\frac{2}{3}}\)

\(\Rightarrow \log x^{x^{2}}=10^{\frac{2}{5}}\)

\(\Rightarrow x^{x^{2}}=10^{10^{\frac{2}{5}}}\)

Comparing the both side, we get \(\mathrm{x}=10\)

We know , the coefficient of middle term in the expansion of \((1+\beta x)^{4}={ }^{4} C_{2} \beta^{2}\)

Similarly the coefficient of middle term in the expansion of \((1-\beta x)^{6}={ }^{6} C_{3}(-\beta)^{3}\)

As it is given \({ }^{4} C_{2} \beta^{2}={ }^{6} C_{3}(-\beta)^{3}\)

\(\Rightarrow { }^{4} C_{2} \beta^{2}={ }^{6} C_{3}(-\beta)^{3} \)

\(\Rightarrow { }^{4} C_{2} \beta^{2}=-{ }^{6} C_{3} \beta^{3} \)

\(\Rightarrow { }^{4} C_{2}+{ }^{6} C_{3} \beta=0\)

\(\Rightarrow \beta=-\frac{{ }^{4} C_{2}}{{ }^{6} C_{3}} \)

\(\Rightarrow \beta=-\frac{\left(\frac{4 !}{2 ! 2 !}\right)}{\left(\frac{6 !}{3 ! 3 !}\right)}\)

\(\Rightarrow \beta=-\frac{\left(\frac{4 \times 3 \times 2 !}{2 \times 2 !}\right)}{\left(\frac{6 \times 5 \times 4 \times 3 !}{3 \times 2 \times 1 \times 3 !}\right)} \)

\(\Rightarrow \beta=-\frac{6}{20} \)

\(\Rightarrow \beta=-\frac{3}{10}\)

Given:

There are 5 subjects and for a student to pass in an examination the students has to pass in each of the 5 subjects.

Case 1: Student fails in any 1 subject out of the 5 subjects.

No. of ways in which student can fail in any 1 subject out of the 5 subjects \(={ }^{5} C_{1}\)

Case 2: Student fails in any 2 subjects out of the 5 subjects.

No. of ways in which student can fail in any 2 subjects out of the 5 subjects \(={ }^{5} C_{2}\)

Case 3: Student fails in any 3 subjects out of the 5 subjects.

No. of ways in which student can fail in any 3 subjects out of the 5 subjects \(={ }^{5} C_{3}\)

Case 4: Student fails in any 4 subjects out of the 5 subjects.

No. of ways in which student can fail in any 4 subjects out of the 5 subjects \(={ }^{5} C_{4}\)

Case 5: Student fails in all 5 subjects.

No. of ways in which student can fail all 5 subjects. \(={ }^{5} C_{5}\)

\(\therefore\) Total number of ways in which a student can fail in an examination \(={ }^{5} C_{1}+{ }^{5} C_{2}+\) \({ }^{5} C_{3}+{ }^{5} C_{4}+{ }^{5} C_{5}=31\)

Given:

\(\mathrm{T}_{\mathrm{r}}\) denotes the \(\mathrm{r}^{\text {th }}\) term in the expansion of \(\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)^{23}\).

We know that, general term expression for \((\mathrm{a}+\mathrm{b})^{\mathrm{n}}\) is:

\(\mathrm{T}_{\mathrm{r}+1}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} \times \mathrm{a}^{\mathrm{n}} \times(\mathrm{b})^{\mathrm{n}-\mathrm{r}}\)

So,

\( \mathrm{T}_{12}={ }^{23} \mathrm{C}_{11} \times \mathrm{x}^{12} \times\left(\frac{1}{\mathrm{x}}\right)^{11}\)

\(\Rightarrow \mathrm{T}_{12}={ }^{23} \mathrm{C}_{11 } x\)

And,

\(\mathrm{T}_{13}={ }^{23} \mathrm{C}_{12} \times \mathrm{x}^{11} \times\left(\frac{1}{\mathrm{x}}\right)^{12}\)

\(\Rightarrow \mathrm{~T}_{13}={ }^{23} \mathrm{C}_{12}\left(\frac{1}{\mathrm{x}}\right)\)

We can write,

\({ }^{23} \mathrm{C}_{12}={ }^{23} \mathrm{C}_{11} \ldots . . \mathrm{as}(11+12=23)\)

\(\Rightarrow \mathrm{T}_{12}={ }^{23} \mathrm{C}_{11} \mathrm{x}^{2} \times \frac{1}{\mathrm{x}}\)

\(\Rightarrow \mathrm{T}_{12}=\mathrm{x}^{2} \times{ }^{23} \mathrm{C}_{12} \times \frac{1}{\mathrm{x}}\)

\(\Rightarrow \mathrm{T}_{12}=\mathrm{x}^{2} . \mathrm{T}_{13}\)

Given:

\(\left(2 x^{2}-\frac{5}{x}\right)^{13}\)

Let \(a= 2x^2,~b=\frac{-5}{x}\)

The general term in the expansion of \((a+b)^n\) is:

\(T_{r+1}={ }^{n} C_{r}\left(a \right)^{n-r} \cdot\left(b\right)^{r}\)

\(T_{r+1}={ }^{13} C_{r}\left(2 x^{2}\right)^{13-r} \cdot\left(-\frac{5}{x}\right)^{r}\)

\(=(-1)^{r} \cdot{ }^{13} C_{r} \cdot 2^{13-r} \cdot 5^{r} \cdot x^{26-3 r} ; \text{so}~ 13 \geq r \geq 0\)

Power of \(x\) in this term is \(26-3 r\).

Total power in all coefficients \(=\sum_{r=0}^{r-13}(26-3 r)\)

\(=26 \sum_{0}^{13} 1-3 \sum_{0}^{13} r \)

\(=26(14)-3\left(\frac{13.14}{2}\right) \)

\(=364-273 \)

\(=91\)

Given expression is \(9^{\text {th }}\) term from the end in \((x-\frac{1 }{ x})^{12}\).

To find \(9^{\text {th }}\) term from the end in \((x-\frac{1 }{ x})^{12}\).

We know that \(r^{\text {th }}\) term from end means \((n-r+2)^{\text {th }}\) term from start.

So, \(9^{\text {th }}\) term from the end \(=[12-9+2]^{\text {th }}\) term from start \(=5^{\text {th }}\) term from start

General term: \(\mathrm{T}_{(\mathrm{r}+1)}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} \times \mathrm{x}^{\mathrm{n}-\mathrm{r}} \times \mathrm{y}^{\mathrm{r}}\)

\(\Rightarrow \mathrm{T}_{5}=\mathrm{T}_{(4+1)}={ }^{12} \mathrm{C}_{4} \times \mathrm{x}^{12-4} \times\left(\frac{-1}{\mathrm{x}}\right)^{4}\)

\(={ }^{12} \mathrm{C}_{4} \times \mathrm{x}^{8} \times \frac{1}{\mathrm{x}^{4}}\)

\(={ }^{12} \mathrm{C}_{4} \mathrm{x}^{4}\)

The \((\mathrm{r}+1)\) th term in the expansion of \(\left(\sqrt{\frac{\mathrm{x}}{3}}-\frac{\sqrt{3}}{\mathrm{x}^{2}}\right)^{10}\) is given by:

\(\mathrm{T}_{\mathrm{r}+1}=(-1)^{\mathrm{r}} \cdot{ }^{10} \mathrm{C}_{\mathrm{r}}\left(\sqrt{\frac{\mathrm{x}}{3}}\right)^{10-\mathrm{r}} \cdot\left(\frac{\sqrt{3}}{\mathrm{x}^{2}}\right)^{\mathrm{r}}\)

\(=(-1)^{\mathrm{r}}{ }^{10} \mathrm{C}_{\mathrm{r}} \mathrm{x}^{5-\frac{5 \mathrm{r}}{2}} 3^{\mathrm{r}-5}\).....(1)

For term to be independent of \(\mathrm{x}\),

\(5-\frac{5 \mathrm{r}}{2}=0\)

\(\Rightarrow \mathrm{r}=2\)

Substituting the values in (1) we get:

\(\mathrm{T}_{3}=(-1)^{2}{ }^{10} \mathrm{C}_{2} 3^{-3}\)

\(=\frac{10(9)}{2(27)}\)

\(=\frac{5}{3}\)

Given:

Expansion \(\left(\frac{a}{b}-\frac{b}{2 a^{2}}\right)^{12}\)

To find the \(9^{\text {th }}\) term in the expansion of \(\left(\frac{a}{b}-\frac{b}{2 a^{2}}\right)^{12}\)

Formula used: (i) \({ }^{n} C_{r}=\frac{n !}{(n-r) !(r) !}\) and \(T_{r+1}={ }^{n} C_{r} a^{n-r} b^{r}\)

For \(9^{\text {th }}\) term,

\(r+1=9\)

\(\Rightarrow r=8 \)

In \(\left(\frac{a}{b}-\frac{b}{2 a^{2}}\right)^{12} \)

\(9^{\text {th }} \text { term }=T_{8+1} \)

\({ }^{12} \mathrm{C}_{8}\left(\frac{a}{b}\right)^{12-8}\left(\frac{-b}{2 a^{2}}\right)^{8} \)

\(\Rightarrow \frac{12 !}{8 !(12-8) !}\left(\frac{a}{b}\right)^{4}\left(\frac{-b}{2 a^{2}}\right)^{8} \)

\(\Rightarrow 495\left(\frac{a^{4}}{b^{4}}\right)\left(\frac{b^{8}}{256 a^{16}}\right) \)

\(\Rightarrow\left(\frac{495 b^{4}}{256 a^{12}}\right) \)