B Given, \(\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)^{2 \mathrm{n}}\) Then, consider \((r+1)^{\text {th }}\) term is independent of \(x\) in the expansion - \(\mathrm{T}_{\mathrm{r}+1} ={ }^{2 \mathrm{n}} \mathrm{C}_{\mathrm{r}}(\mathrm{x})^{2 \mathrm{n}-\mathrm{r}}\left(\frac{1}{\mathrm{x}}\right)^{\mathrm{r}}\) \(={ }^{2 \mathrm{n}} \mathrm{C}_{\mathrm{r}} \mathrm{x}^{2 \mathrm{n}-\mathrm{r}}\left(\mathrm{x}^{-1}\right)^{\mathrm{r}}\) \(\mathrm{T}_{\mathrm{r}+1} ={ }^{2 \mathrm{n}} \mathrm{C}_{\mathrm{r}} \mathrm{x}^{2 \mathrm{n}-2 \mathrm{r}}\) The term is independent of \(x\) then power of \(x\) should be zero \(2 \mathrm{n}-2 \mathrm{r}=0\) \(\mathrm{n}=\mathrm{r}\) So, \({ }^{2 \mathrm{n}} \mathrm{C}_{\mathrm{n}}=\frac{2 \mathrm{n} !}{(2 \mathrm{n}-\mathrm{n}) ! \mathrm{n} !}=\frac{1 \cdot 2 \cdot 3 \cdot 4 \ldots \ldots(2 \mathrm{n}-1)(2 \mathrm{n})}{\mathrm{n} ! \mathrm{n} !}\) \(=\frac{(2 n-1)(2 n-3)(2 n-5) \ldots . .3 .2\left(2^{\mathrm{n}}\right)}{n !}\) \(=\frac{1.3 .5 \ldots(2 \mathrm{n}-1) \times 2^{\mathrm{n}}}{n !}\)
JCECE-2013
Binomial Theorem and its Simple Application
119337
Let \(T_n\) denote the number of triangles which can be formed using the vertices of a regular polygon of \(n\) sides. If \(T_{n+1}-T_n=21\), then \(n\) equals:
1 5
2 7
3 6
4 4
Explanation:
B Given, the \(T_n\) denotes the number of triangle and which has the side \(n\). And, \(\mathrm{T}_{\mathrm{n}+1}-\mathrm{T}_{\mathrm{n}}=21\) Then, \({ }^{n+1} C_3-{ }^n C_3=21\) \(\frac{(\mathrm{n}+1) !}{3 !(\mathrm{n}+1-3) !}-\frac{\mathrm{n} !}{3 !(\mathrm{n}-3) !}=21\) \(\frac{(\mathrm{n}+1) \mathrm{n} \cdot(\mathrm{n}-1)(\mathrm{n}-2) !}{3 \times 2 \times(\mathrm{n}-2) !}-\frac{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)(\mathrm{n}-3) !}{3 \times 2 \times(\mathrm{n}-3) !}=21\) \(\frac{\mathrm{n}(\mathrm{n}+1)(\mathrm{n}-1)}{6}-\frac{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)}{6}=21\) \(\mathrm{n}(\mathrm{n}+1)(\mathrm{n}-1)-(\mathrm{n})(\mathrm{n}-1)(\mathrm{n}-2)=21 \times 6\) \(\left(\mathrm{n}^2+\mathrm{n}\right)(\mathrm{n}-1)-\left(\mathrm{n}^2-\mathrm{n}\right)(\mathrm{n}-2)=21 \times 6\) \(\left(n^3+n^2-n^2-n\right)-\left(n^3-n^2-2 n^2+2 n\right)=126\) \(n^3+n^2-n^2-n-n^3+n^2+2 n^2-2 n=126\) \(3 n^2-3 n=126\) \(\mathrm{n}^2-\mathrm{n}=42\) \(\mathrm{n}^2-\mathrm{n}-42=0\) \(\mathrm{n}(\mathrm{n}-7)+6(\mathrm{n}-7)=0\) \(n^2-7 n+6 n-42=0\) \((n-7)(n+6)=0\) \(\mathrm{n}=7(+\mathrm{ve})\) So, \(n\) equals to 7 .
JCECE-2006
Binomial Theorem and its Simple Application
119338
Let \((1+x)^{36}=a_0+a_1 x+a_2 x^2+\ldots+a_{36} x^{36}\), Then, \(a_0+a_3+a_6+\ldots+a_{36}\) is equal to
B Given, \(\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)^{2 \mathrm{n}}\) Then, consider \((r+1)^{\text {th }}\) term is independent of \(x\) in the expansion - \(\mathrm{T}_{\mathrm{r}+1} ={ }^{2 \mathrm{n}} \mathrm{C}_{\mathrm{r}}(\mathrm{x})^{2 \mathrm{n}-\mathrm{r}}\left(\frac{1}{\mathrm{x}}\right)^{\mathrm{r}}\) \(={ }^{2 \mathrm{n}} \mathrm{C}_{\mathrm{r}} \mathrm{x}^{2 \mathrm{n}-\mathrm{r}}\left(\mathrm{x}^{-1}\right)^{\mathrm{r}}\) \(\mathrm{T}_{\mathrm{r}+1} ={ }^{2 \mathrm{n}} \mathrm{C}_{\mathrm{r}} \mathrm{x}^{2 \mathrm{n}-2 \mathrm{r}}\) The term is independent of \(x\) then power of \(x\) should be zero \(2 \mathrm{n}-2 \mathrm{r}=0\) \(\mathrm{n}=\mathrm{r}\) So, \({ }^{2 \mathrm{n}} \mathrm{C}_{\mathrm{n}}=\frac{2 \mathrm{n} !}{(2 \mathrm{n}-\mathrm{n}) ! \mathrm{n} !}=\frac{1 \cdot 2 \cdot 3 \cdot 4 \ldots \ldots(2 \mathrm{n}-1)(2 \mathrm{n})}{\mathrm{n} ! \mathrm{n} !}\) \(=\frac{(2 n-1)(2 n-3)(2 n-5) \ldots . .3 .2\left(2^{\mathrm{n}}\right)}{n !}\) \(=\frac{1.3 .5 \ldots(2 \mathrm{n}-1) \times 2^{\mathrm{n}}}{n !}\)
JCECE-2013
Binomial Theorem and its Simple Application
119337
Let \(T_n\) denote the number of triangles which can be formed using the vertices of a regular polygon of \(n\) sides. If \(T_{n+1}-T_n=21\), then \(n\) equals:
1 5
2 7
3 6
4 4
Explanation:
B Given, the \(T_n\) denotes the number of triangle and which has the side \(n\). And, \(\mathrm{T}_{\mathrm{n}+1}-\mathrm{T}_{\mathrm{n}}=21\) Then, \({ }^{n+1} C_3-{ }^n C_3=21\) \(\frac{(\mathrm{n}+1) !}{3 !(\mathrm{n}+1-3) !}-\frac{\mathrm{n} !}{3 !(\mathrm{n}-3) !}=21\) \(\frac{(\mathrm{n}+1) \mathrm{n} \cdot(\mathrm{n}-1)(\mathrm{n}-2) !}{3 \times 2 \times(\mathrm{n}-2) !}-\frac{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)(\mathrm{n}-3) !}{3 \times 2 \times(\mathrm{n}-3) !}=21\) \(\frac{\mathrm{n}(\mathrm{n}+1)(\mathrm{n}-1)}{6}-\frac{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)}{6}=21\) \(\mathrm{n}(\mathrm{n}+1)(\mathrm{n}-1)-(\mathrm{n})(\mathrm{n}-1)(\mathrm{n}-2)=21 \times 6\) \(\left(\mathrm{n}^2+\mathrm{n}\right)(\mathrm{n}-1)-\left(\mathrm{n}^2-\mathrm{n}\right)(\mathrm{n}-2)=21 \times 6\) \(\left(n^3+n^2-n^2-n\right)-\left(n^3-n^2-2 n^2+2 n\right)=126\) \(n^3+n^2-n^2-n-n^3+n^2+2 n^2-2 n=126\) \(3 n^2-3 n=126\) \(\mathrm{n}^2-\mathrm{n}=42\) \(\mathrm{n}^2-\mathrm{n}-42=0\) \(\mathrm{n}(\mathrm{n}-7)+6(\mathrm{n}-7)=0\) \(n^2-7 n+6 n-42=0\) \((n-7)(n+6)=0\) \(\mathrm{n}=7(+\mathrm{ve})\) So, \(n\) equals to 7 .
JCECE-2006
Binomial Theorem and its Simple Application
119338
Let \((1+x)^{36}=a_0+a_1 x+a_2 x^2+\ldots+a_{36} x^{36}\), Then, \(a_0+a_3+a_6+\ldots+a_{36}\) is equal to
B Given, \(\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)^{2 \mathrm{n}}\) Then, consider \((r+1)^{\text {th }}\) term is independent of \(x\) in the expansion - \(\mathrm{T}_{\mathrm{r}+1} ={ }^{2 \mathrm{n}} \mathrm{C}_{\mathrm{r}}(\mathrm{x})^{2 \mathrm{n}-\mathrm{r}}\left(\frac{1}{\mathrm{x}}\right)^{\mathrm{r}}\) \(={ }^{2 \mathrm{n}} \mathrm{C}_{\mathrm{r}} \mathrm{x}^{2 \mathrm{n}-\mathrm{r}}\left(\mathrm{x}^{-1}\right)^{\mathrm{r}}\) \(\mathrm{T}_{\mathrm{r}+1} ={ }^{2 \mathrm{n}} \mathrm{C}_{\mathrm{r}} \mathrm{x}^{2 \mathrm{n}-2 \mathrm{r}}\) The term is independent of \(x\) then power of \(x\) should be zero \(2 \mathrm{n}-2 \mathrm{r}=0\) \(\mathrm{n}=\mathrm{r}\) So, \({ }^{2 \mathrm{n}} \mathrm{C}_{\mathrm{n}}=\frac{2 \mathrm{n} !}{(2 \mathrm{n}-\mathrm{n}) ! \mathrm{n} !}=\frac{1 \cdot 2 \cdot 3 \cdot 4 \ldots \ldots(2 \mathrm{n}-1)(2 \mathrm{n})}{\mathrm{n} ! \mathrm{n} !}\) \(=\frac{(2 n-1)(2 n-3)(2 n-5) \ldots . .3 .2\left(2^{\mathrm{n}}\right)}{n !}\) \(=\frac{1.3 .5 \ldots(2 \mathrm{n}-1) \times 2^{\mathrm{n}}}{n !}\)
JCECE-2013
Binomial Theorem and its Simple Application
119337
Let \(T_n\) denote the number of triangles which can be formed using the vertices of a regular polygon of \(n\) sides. If \(T_{n+1}-T_n=21\), then \(n\) equals:
1 5
2 7
3 6
4 4
Explanation:
B Given, the \(T_n\) denotes the number of triangle and which has the side \(n\). And, \(\mathrm{T}_{\mathrm{n}+1}-\mathrm{T}_{\mathrm{n}}=21\) Then, \({ }^{n+1} C_3-{ }^n C_3=21\) \(\frac{(\mathrm{n}+1) !}{3 !(\mathrm{n}+1-3) !}-\frac{\mathrm{n} !}{3 !(\mathrm{n}-3) !}=21\) \(\frac{(\mathrm{n}+1) \mathrm{n} \cdot(\mathrm{n}-1)(\mathrm{n}-2) !}{3 \times 2 \times(\mathrm{n}-2) !}-\frac{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)(\mathrm{n}-3) !}{3 \times 2 \times(\mathrm{n}-3) !}=21\) \(\frac{\mathrm{n}(\mathrm{n}+1)(\mathrm{n}-1)}{6}-\frac{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)}{6}=21\) \(\mathrm{n}(\mathrm{n}+1)(\mathrm{n}-1)-(\mathrm{n})(\mathrm{n}-1)(\mathrm{n}-2)=21 \times 6\) \(\left(\mathrm{n}^2+\mathrm{n}\right)(\mathrm{n}-1)-\left(\mathrm{n}^2-\mathrm{n}\right)(\mathrm{n}-2)=21 \times 6\) \(\left(n^3+n^2-n^2-n\right)-\left(n^3-n^2-2 n^2+2 n\right)=126\) \(n^3+n^2-n^2-n-n^3+n^2+2 n^2-2 n=126\) \(3 n^2-3 n=126\) \(\mathrm{n}^2-\mathrm{n}=42\) \(\mathrm{n}^2-\mathrm{n}-42=0\) \(\mathrm{n}(\mathrm{n}-7)+6(\mathrm{n}-7)=0\) \(n^2-7 n+6 n-42=0\) \((n-7)(n+6)=0\) \(\mathrm{n}=7(+\mathrm{ve})\) So, \(n\) equals to 7 .
JCECE-2006
Binomial Theorem and its Simple Application
119338
Let \((1+x)^{36}=a_0+a_1 x+a_2 x^2+\ldots+a_{36} x^{36}\), Then, \(a_0+a_3+a_6+\ldots+a_{36}\) is equal to
B Given, \(\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)^{2 \mathrm{n}}\) Then, consider \((r+1)^{\text {th }}\) term is independent of \(x\) in the expansion - \(\mathrm{T}_{\mathrm{r}+1} ={ }^{2 \mathrm{n}} \mathrm{C}_{\mathrm{r}}(\mathrm{x})^{2 \mathrm{n}-\mathrm{r}}\left(\frac{1}{\mathrm{x}}\right)^{\mathrm{r}}\) \(={ }^{2 \mathrm{n}} \mathrm{C}_{\mathrm{r}} \mathrm{x}^{2 \mathrm{n}-\mathrm{r}}\left(\mathrm{x}^{-1}\right)^{\mathrm{r}}\) \(\mathrm{T}_{\mathrm{r}+1} ={ }^{2 \mathrm{n}} \mathrm{C}_{\mathrm{r}} \mathrm{x}^{2 \mathrm{n}-2 \mathrm{r}}\) The term is independent of \(x\) then power of \(x\) should be zero \(2 \mathrm{n}-2 \mathrm{r}=0\) \(\mathrm{n}=\mathrm{r}\) So, \({ }^{2 \mathrm{n}} \mathrm{C}_{\mathrm{n}}=\frac{2 \mathrm{n} !}{(2 \mathrm{n}-\mathrm{n}) ! \mathrm{n} !}=\frac{1 \cdot 2 \cdot 3 \cdot 4 \ldots \ldots(2 \mathrm{n}-1)(2 \mathrm{n})}{\mathrm{n} ! \mathrm{n} !}\) \(=\frac{(2 n-1)(2 n-3)(2 n-5) \ldots . .3 .2\left(2^{\mathrm{n}}\right)}{n !}\) \(=\frac{1.3 .5 \ldots(2 \mathrm{n}-1) \times 2^{\mathrm{n}}}{n !}\)
JCECE-2013
Binomial Theorem and its Simple Application
119337
Let \(T_n\) denote the number of triangles which can be formed using the vertices of a regular polygon of \(n\) sides. If \(T_{n+1}-T_n=21\), then \(n\) equals:
1 5
2 7
3 6
4 4
Explanation:
B Given, the \(T_n\) denotes the number of triangle and which has the side \(n\). And, \(\mathrm{T}_{\mathrm{n}+1}-\mathrm{T}_{\mathrm{n}}=21\) Then, \({ }^{n+1} C_3-{ }^n C_3=21\) \(\frac{(\mathrm{n}+1) !}{3 !(\mathrm{n}+1-3) !}-\frac{\mathrm{n} !}{3 !(\mathrm{n}-3) !}=21\) \(\frac{(\mathrm{n}+1) \mathrm{n} \cdot(\mathrm{n}-1)(\mathrm{n}-2) !}{3 \times 2 \times(\mathrm{n}-2) !}-\frac{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)(\mathrm{n}-3) !}{3 \times 2 \times(\mathrm{n}-3) !}=21\) \(\frac{\mathrm{n}(\mathrm{n}+1)(\mathrm{n}-1)}{6}-\frac{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)}{6}=21\) \(\mathrm{n}(\mathrm{n}+1)(\mathrm{n}-1)-(\mathrm{n})(\mathrm{n}-1)(\mathrm{n}-2)=21 \times 6\) \(\left(\mathrm{n}^2+\mathrm{n}\right)(\mathrm{n}-1)-\left(\mathrm{n}^2-\mathrm{n}\right)(\mathrm{n}-2)=21 \times 6\) \(\left(n^3+n^2-n^2-n\right)-\left(n^3-n^2-2 n^2+2 n\right)=126\) \(n^3+n^2-n^2-n-n^3+n^2+2 n^2-2 n=126\) \(3 n^2-3 n=126\) \(\mathrm{n}^2-\mathrm{n}=42\) \(\mathrm{n}^2-\mathrm{n}-42=0\) \(\mathrm{n}(\mathrm{n}-7)+6(\mathrm{n}-7)=0\) \(n^2-7 n+6 n-42=0\) \((n-7)(n+6)=0\) \(\mathrm{n}=7(+\mathrm{ve})\) So, \(n\) equals to 7 .
JCECE-2006
Binomial Theorem and its Simple Application
119338
Let \((1+x)^{36}=a_0+a_1 x+a_2 x^2+\ldots+a_{36} x^{36}\), Then, \(a_0+a_3+a_6+\ldots+a_{36}\) is equal to
