Binomial Theorem Test - 79

Given:

\(\left(x^{2}-\frac{1}{x}\right)^{9}\)

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

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

\(\Rightarrow \mathrm{T}_{(\mathrm{r}+1)}={ }^{9} \mathrm{C}_{\mathrm{r}} \times\left(\mathrm{x}^{2}\right)^{9-\mathrm{r}} \times\left(\frac{-1}{\mathrm{x}}\right)^{\mathrm{r}}\)

\(={ }^{9} \mathrm{C}_{r}\cdot \mathrm{x}^{18-2 \mathrm{r}}(-1)^{\mathrm{r}} x^{-r}\)

\(=(-1)^{r}\cdot{ }^{9} \mathrm{C}_{r} \mathrm{x}^{18-3 r}\)

For the term independent of \(x\), power of \(x\) should be zero

Therefore, \(18-3 r=0\)

\(\therefore \mathrm{r}=6\)

So, the value is \((-1)^{6}\cdot{ }^{9} \mathrm{C}_{6}=\frac{9!}{3!\times 6!}\)

\(=\frac{7×8×9}{3×2×1}\)

\(=\frac{504}{6}=84\)

The expression \(\left(\frac{1+x}{1-x}\right)^{2}\) can be written as:

\(=(1+x)^{2} \times(1-x)^{-2} \)

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

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

\(x^{n}\) will be obtained by multiplying \(\left(1\right.\) and \(\left.{ }^{n+1} C_{n} x^{n}\right)\) and \(\left(2 x\right.\) and \(\left.{ }^{n} C_{n-1} x^{n-1}\right)\) and \(\left(x^{2}\right.\) and \({ }^{n-1} C_{n-2} x^{n-}\) \({ }^{2}\) ).

\(\therefore\) Coefficient of \(x^{n}\) will be: \(\left(1 \times{ }^{n+1} C_{n}\right)+\left(2 \times{ }^{n} C_{n-1}\right)+\left(1 \times{ }^{n-1} C_{n-2}\right)\)

\(={ }^{n+1} C_{n}+2 \times{ }^{n} C_{n-1}+{ }^{n-1} C_{n-2}\)

Given:

The binomial coefficients of \((3 r-1)\) and \((r+2)\) th terms are equal.

3rd term in the expansion of \((1+x)^{2 n}\)

\(={ }^{2 \mathrm{n}} \mathrm{C}_{3 \mathrm{r}-1} \mathrm{x}^{3 \mathrm{r}-1}\)

\((r+2)\) th term in the expansion of \((1+x)\)

\(={ }^{2 \mathrm{n}} \mathrm{C}_{\mathrm{r}+1} \mathrm{x}^{\mathrm{r}+1}\)

According to question:

\({ }^{2 \mathrm{n}} \mathrm{C}_{3 \mathrm{r}-1}={ }^{2 \mathrm{n}} \mathrm{C}_{\mathrm{r}+1}\)

\(\Rightarrow 3 \mathrm{r}-1=\mathrm{r}+1\)

So, \(2 \mathrm{n}=(3 \mathrm{r}-1)+(\mathrm{r}+1)\)

\( 2 \mathrm{r}=2\) or \(2 \mathrm{n}=4 \mathrm{r}\)

\(\Rightarrow \mathrm{r}=1\) or \(\mathrm{n}=2 \mathrm{r}\)

But \(\mathrm{r}>1\)

Therefore, \(\mathrm{n}=2 \mathrm{r}\)

Given:

\((x+a)^{47}-(x-a)^{47}\)

When we expand the above equation using binomial expansion

\((\mathrm{x}+\mathrm{y})^{\mathrm{n}}=\left(\sum_{\mathrm{k}=0}^{\mathrm{n}}{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{k}} \mathrm{x}^{\mathrm{k}} \mathrm{y}^{\mathrm{n}-\mathrm{k}}\right)\)

So the above equation becomes

\((\mathrm{x}+\mathrm{a})^{47}=\left(\sum_{\mathrm{k}=0}^{47}{ }^{47} \mathrm{C}_{\mathrm{k}} \mathrm{x}^{\mathrm{k}} \mathrm{a}^{47-\mathrm{k}}\right) \)

\((\mathrm{x}-\mathrm{a})^{47}=\left(\sum_{\mathrm{k}=0}^{47}{ }^{47} \mathrm{C}_{\mathrm{k}} \mathrm{x}^{\mathrm{k}}(-\mathrm{a})^{47-\mathrm{k}}\right)\)

\((x+a)^{47} \rightarrow\) There are 48 terms in the expansion and all are positive.

\((x-a)^{47} \rightarrow\) There are 48 terms in the expansion.

The terms with odd powers of a will be cancelled and those with even powers of a will add up.

24 terms will be positive and 24 negative in the expansion of \((x-a)^{47}\)

48 terms positive- [ 24 terms negative and 24 terms positive]

\(=48\) terms positive \(+24\) terms negative \(+24\) terms positive \(=24\) terms

Given expansion \(\left(\mathrm{y}^{\frac{1}{5}}+\mathrm{x}^{\frac{1}{10}}\right)^{55}\)

In the expansion of \(\left(\mathrm{y}^{\frac{1 }{ 5}}+\mathrm{x}^{\frac{1 }{ 10}}\right)^{55}\), the general term is:

\(\mathrm{T}_{\mathrm{r}+1}={ }^{55} \mathrm{C}_{\mathrm{r}}\left(\mathrm{y}^{\frac{1}{ 5}}\right)^{55-\mathrm{r}}\left(\mathrm{x}^{\frac{1 }{ 10}}\right)^{\mathrm{r}}\)

\(={ }^{55} \mathrm{C}_{\mathrm{r}} \cdot \mathrm{y}^{11-\frac{\mathrm{r} }{ 5}} \mathrm{x}^{\frac{\mathrm{r} }{ 10}}\)

This \(T_{\mathrm{r}+1}\) will be independent of radicals if the exponents \(\frac{\mathrm{r} }{ 5}\) and \(\frac{\mathrm{r} }{ 10} \) are integers, for \(0 \leq \mathrm{r} \leq 55\) which is possible only when \(\mathrm{r}=0,10,20,30,40,50\).

\(\therefore\) There are six terms viz. \(\mathrm{T}_{1}, \mathrm{~T}_{11}, \mathrm{~T}_{21}, \mathrm{~T}_{31}, \mathrm{~T}_{41}, \mathrm{~T}_{51}\), which are independent of radicals.

Given expression is \((2 \mathrm{x}+3 \mathrm{y}+\mathrm{z})^{7}\).

Number of dissimilar terms in the expansion of \(\left(\mathrm{x}_{1}+\mathrm{x}_{2}+\mathrm{x}_{3}+\ldots \ldots+\right.\) \(\left.\mathrm{x}_{\mathrm{r}}\right)^{\mathrm{n}}\) is \({ }^{\mathrm{n}+\mathrm{r}-1} \mathrm{C}_{\mathrm{n}}\).

So in \((2 x+3 y+z)^{7}\), we have \(r=3\) and \(n=7\)

Thus number of terms in the expansion of \((2 x+3 y+z)^{7}\) is \(^{7+3-1} C_{7}={ }^{9} C_{7}\)

\(=\) 36

Given for the expression \(\left(x^{2}+\frac{\lambda}{x}\right)^{5}, n=5\)

General term in the expansion of \((a+b)^{n}\) is given by:

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

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

\(={ }^{5} C_{r} \lambda^{r} x^{10-3 r}\)

For the term \(x^{}, \)

\(10-3 r=1 \)

\(\Rightarrow r=3\)

\(\therefore{ }^{5} \mathrm{C}_{3} \cdot\lambda^{3}=270\)

\(\Rightarrow \frac{5!}{2!×3!} \cdot\lambda^{3}=270\)

\(\Rightarrow \frac{5×4×3×2×1}{2×1×3×2×1} \cdot\lambda^{3}=270\)

\(\Rightarrow \frac{5×4}{2×1} \cdot\lambda^{3}=270\)

\(\Rightarrow 10 \cdot\lambda^{3}=270\)

\(\Rightarrow \lambda^{3}=27\)

\(\Rightarrow \lambda=3\)

Given:

\(\left(1+3 x+3 x^{2}+x^{3}\right)^{2 n}\)

\(=(1+x)^{6 n}\)

Here, \(n\) is an even number.

So the middle terms is \((\frac{6 n}{2} { +1})=(3 n+1)^{\text {th }}\) term

So terms \((3 n+1)^{\text {th }}\) term is:

\(T_{3 n+1}={ }^{6 n} C_{3 n} x^{3 n} \)

\(=\frac{(6 n) !}{(3 n!)^{2}} x^{3 n}\)

Given expression is \(\left(x-\frac{1}{x}\right)^{10}\).

General term in the expansion of \((a+b)^{n}\) is given by:

\(T_{(r+1)}={ }^{n} C_{,} \times a^{n-7} \times b^{r}\)

In the given binomial expression \(\left(x-\frac{1}{x}\right)^{10}, n=10, a=x\) and \(b=\frac{-1}{x}\)

\(\therefore T_{r-1}={ }^{10} C_{r} x^{10-r}\left(\frac{-1}{x}\right)^{r}\)

\(={ }^{10} C_{n}(-1)^{r} x^{10-2 r}\)

For the term to be independent of \(x\), we must have

\(10-2 r=0\)

\(\Rightarrow r=5\)

The required term is \({ }^{10} C_{5}(-1)^{5}=-{ }^{10} C_{5}\)

Given:

The second term in the expansion of \((2+\mathrm{x})^{5}\) is 240.

We know that:

The general term of \((\mathrm{a}+\mathrm{x})^{\mathrm{n}}\) is \({ }^{\mathrm{n}} \mathrm{C}_ \mathrm{r} \mathrm{a}^{(\mathrm{n}-\mathrm{r})} \mathrm{x}^{\mathrm{r}}\)

Here, \(\mathrm{n}=5\)

we will get second term if we put \(\mathrm{r}=1\)

The second term will be \({ }^{5} \mathrm{C}_{1}\left(2^{(5-1)}\right) \mathrm{x}\) \(=5\left(2^{4}\right) \mathrm{x}\)

According to the question:

\(5\left(2^{4}\right) \mathrm{x}=240\)

Therefore

\(5(16) \mathrm{x}=240 \)

\(\Rightarrow 80(\mathrm{x})=240 \)

\(\Rightarrow \mathrm{x}=3\)