Binomial Theorem Test - 78

Given expression is\((8 x+2 y)^{23}\).

The formula for the general term of the binomial expansion of \((a+b)^{n}\) is:

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

Therefore, we can write the general term of the expansion of \((8 x+2 y)^{23}\) by setting \(a=8 x, b=2 y\), and \(n=23\) as follows:

\(T_{r+1}={ }^{23} C_{r}(8 x)^{23-r}(2 y)^{r}\)

\(={ }^{23} C_{r} 8^{23-r} x^{23-r} 2^{r} y^{r}\)

Therefore, the ratio between the eighth and seventh term is given by:

\(\frac{T_{8}}{T_{7}}=\frac{{ }^{23} C_{7} 8^{23-7} 2^{7} x^{23-7} y^{7}}{{ }^{23} C_{6} 8^{23-6} 2^{6} x^{23-6} y^{6}}\)

Using the rules of exponents, we can simplify this to:

\(\frac{T_{8}}{T_{7}}=\frac{{ }^{23} C_{7} 2 y}{{ }^{23} C_{6} 8 x}\)

\(=\frac{{ }^{23} C_{7} y}{{ }^{23} C_{6} 4 x}\)

The ratio of consecutive combinations is given by:

\(\frac{{ }^{n} C_{r}}{{ }^{n} C_{r-1}}=\frac{n-r+1}{r}\)

\(\frac{{ }^{23} C_{7}}{{ }^{23} C_{6}}=\frac{23-7+1}{7}=\frac{17}{7}\)

Substituting this into the equation above, we have

\(\frac{T_{8}}{T_{7}}=\frac{17 y}{7 \times 4 x}\)

\(=\frac{17 y}{28 x}\)

Let \(a_{1}, a_{2}, a_{3}\) and \(a_{4}\) be the coefficients of 4 consecutive terms viz. the rth, \((r+1)\) th, \((r+2)\) th and \((r+3)\) th terms.

Then, \(a_{1}={ }^{n} C_{r-1}, a_{2}=" C_{r}, a_{3}={ }^{n} C_{r+1}\) and \(a_{4}={ }^{n} C_{t-2}\). Now,

\(\frac{a_{1}}{a_{1}+a_{2}}+\frac{a_{3}}{a_{3}+a_{4}}=\frac{1}{1+\frac{a_{2}}{a_{1}}}+\frac{1}{1+\frac{a_{4}}{a_{3}}}\)

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

\(=\frac{1}{1+\frac{n-r+1}{r}}+\frac{1}{1+\frac{n-r-1}{r+2}}\)

\(=\frac{r}{n+1}+\frac{r+2}{n+1}\)

\(=\frac{2 \cdot(r+1)}{n+1}\)

Given:

\(\left(x^{3}-\frac{1}{x^{2}}\right)^{15}\)

Let \((r+1)^{t h}\) term be the constant term in the expansion of \(\left(x^{3}-\frac{1}{x^{2}}\right)^{15}\).

We know that in the binomial expansion of \((a+x)^{n}\), we have,

\(T_{r+1}={ }^{n} C_{r} x^{n-r} a^{r}\)

\(\therefore T_{r+1}={ }^{15} C_{r}\left(x^{3}\right)^{15-r}\left(-\frac{1}{x^{2}}\right)^{r}\)

\(T_{r+1}={ }^{15} C_{r} x^{45-5 r}(-1)^{r}\) is independent of \(x\). If:

\(45-5 r=0\)

\(\Rightarrow r=9\)

Thus, \(10^{\text {th }}\) term is independent of \(x\) and is given by

\(T_{10}={ }^{15} C_{9}(-1)^{9}\)

\(=-{ }^{15} C_{9}\)

Given:\(\frac{x^{3 / 2}-1}{x+1+x^{12}}-\frac{x^{32}+1}{x+1-x^{12}}\)

\(=\frac{\left(x^{12}\right)^{3}-1}{\left(x^{12}\right)^{2}+x^{12}+1}-\frac{\left(x^{1 / 2}\right)^{3}+1}{\left(x^{12}\right)^{2}-x^{12}+1} \)

\(=\frac{\left(x^{12}-1\right)\left\{\left(x^{12}\right)^{2}+x^{1 / 2}+1\right\}}{\left(x^{12}\right)^{2}+x^{12}+1}-\frac{\left(x^{1 / 2}+1\right)\left\{\left(x^{12}\right)^{2}-x^{12}+1\right\}}{\left(x^{12}\right)^{2}-x^{12}+1} \)

\(=\left(x^{12}-1\right)-\left(x^{12}+1\right) \)\(=-2\)

So, \(\left(\frac{x^{3 / 2}-1}{x+1+x^{1 / 2}}-\frac{x^{3 / 2}+1}{x+1-x^{1 / 2}}\right)^{5}=(-2)^{5}\)\(=-32\)

So, the expansion has only one term which is \(-32\) and independent of \(x\).

Given:

\({ }^{n} C_{0}=1\)

\({ }^{n} C_{1}(a x)=12 x \).....(1)

\({ }^{n} C_{2}(a x)^{2}=64 x^{2}\).....(2)

The first three terms in the expansion of \((1+a x)^{n}\) are \({ }^{n} C_{0},{ }^{n} C_{1}(a x),{ }^{n} C_{2}(a x)^{2}\).

From (1)

\({ }^{n} C_{1}(a x)=12 x \)

\(\Rightarrow \frac{n ! a}{11 \cdot(n-1) !}=12\)

\(\Rightarrow n a=12\)

\(\Rightarrow a=\frac{12}{n}\).....(3)

From (2)

\({ }^{n} C_{2}(a x)^{2}=64 x^{2}\)

\(\Rightarrow { }^{n} C_{2} \cdot(a)^{2}-64\)

\(\Rightarrow \frac{n !}{(n-2) ! 21}\left(a^{2}\right)=64\)

\(\Rightarrow \frac{n \cdot(n-1)\left(a^{2}\right)}{2}=64\).....(4)

Put the value from (3) into (4).

\(\frac{n(n-1)}{2} \times\left(\frac{144}{n^{2}}\right)=64 \)

\(\Rightarrow \frac{n-1}{n}=\frac{64}{72}=\frac{8}{9}\)

\(\Rightarrow 9 n-9=8 n\)

\(\Rightarrow n=9\)

Given:

\(\left(1+x-2 x^{2}\right)^{6}=1+a_{1} x+a_{2} x^{2}+\ldots \ldots+a_{12} x^{12}\)

Put \(\mathrm{x}=1\) in the given equation

\(0=1+\mathrm{a}_{1}+\mathrm{a}_{2}+\mathrm{a}_{3}+\ldots .+\mathrm{a}_{12}\).....(1)

Put \(\mathrm{x}=-1\) in the given equation

\((-2)^{6}=1-\mathrm{a}_{1}+\mathrm{a}_{2}-\mathrm{a}_{3}+\ldots+\mathrm{a}_{12}\).....(2)

Adding (1) and (2), we get

\(0+(-2)^{6}=\left(1+\mathrm{a}_{1}+\mathrm{a}_{2}+\ldots+\mathrm{a}_{12}\right)+\left(1-\mathrm{a}_{1}+\mathrm{a}_{2}-\mathrm{a}_{3} \ldots \ldots+\mathrm{a}_{12}\right) \)

\(\Rightarrow 64=2\left(1+\mathrm{a}_{2}+\mathrm{a}_{4}+\ldots .+\mathrm{a}_{12}\right) \)

\(\Rightarrow 1+\mathrm{a}_{2}+\mathrm{a}_{4}+\ldots \ldots+\mathrm{a}_{12}=32 \)

\(\Rightarrow \mathrm{a}_{2}+\mathrm{a}_{4}+\ldots \ldots+\mathrm{a}_{12}=31\)

Given:

\(\left(1+3 x+3 x^{2}+x^{3}\right)^{6}\)

\(=\left[(1+\mathrm{x})^{3}\right]^{6} \)

\(=[1+\mathrm{x}]^{6(3)} \)

\(=(1+\mathrm{x})^{18}\)

So, total number of terms will be \(18+1=19\) terms.

Given expression is \((a+b)^{n}\).

The coefficients of \(3^{\text {rd }}\) and \(15^{\text {th }}\) terms in the binomial expansion of \((a+b)^{n}\) are \({ }^{n} C_{2}\) and \({ }^{n} C_{14}\) respectively. It is given that:

Coefficient of \(3^{\text {rd }}\) term in \((a+b)^{n}=\) coefficient of \(15^{\text {th }}\) term in \((a+b)^{n}\)

\({ }^{n} C_{2} ={ }^{n} C_{14} \) \(\left[\because{ }^{n} C_{x}={ }^{n} C_{y} \rightarrow x=y, \text { or } x+y=n\right]\)

So,

\(n =2+14 \)

\(=16 \)

Given:

\(\left(1+x+x^{2}+x^{3}\right)^{11}\).

By expanding given equation using expansion formula we can get the coefficient \(x^{4}\).

\( 1+x+x^{2}+x^{3}\)

\(\Rightarrow(1+x)+x^{2}(1+x)\)

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

So,

\(\left(1+x+x^{2}+x^{3}\right)^{11}=(1+x)^{11}\left(1+x^{2}\right)^{11}\)

\(=1+{ }^{11} C_{1} x^{2}+{ }^{11} C_{2} x^{2}+{ }^{11} C_{3} x^{3}+{ }^{11} C_{4} x^{4} \ldots \ldots\)

\(=1+{ }^{11} C_{1} x^{2}+{ }^{11} C_{2} x^{4}+\ldots \ldots\)

To find term in from the product of two brackets on the right-hand-side, consider the following products terms as,

\(=1 \times{ }^{11} C_{2} x^{4}+{ }^{11} C_{2} x^{2} \times{ }^{11} C_{1} x^{2}+{ }^{11} C_{4} x^{4}\)

\(\Rightarrow\left[{ }^{11} C_{2}+{ }^{11} C_{2} \times{ }^{11} C_{1}+{ }^{11} C_{4}\right] x^{4}\)

\(\Rightarrow[55+605+330] x^{4}\)

\(\Rightarrow 990 x^{4}\)

So,

The coefficient of \(x^{4}\) is 990.

Given:

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

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

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

So,

\(T_{r+1}={ }^{n} C_{r}\left(x^{2}\right)^{n-7}\left(\frac{1}{x^{3}}\right)^{r}\)

\(\Rightarrow T_{r+1}={ }^{n} C_{r} x^{2 n-5 r}\)

So, \(2 n-5 r=1\).....(1)

According to question:

\({ }^{n} C_{r}={ }^{n} C_{23}\)

so, \(r=23\) or \(n-r=23 \quad \ldots\)(2)

From (1) and (2),

Minimum value is \(n=38\)