D : Given, \((\sqrt{3}+\sqrt{2})^4-(\sqrt{3}-\sqrt{2})^4\) \((x+y)^4-(x-y)^4=8 x^3 y+8 x y^3\) \(=8 x y\left(x^2+y^2\right)\) \(x=\sqrt{3}, y=\sqrt{2}\) \((\sqrt{3}+\sqrt{2})^4-(\sqrt{3}-\sqrt{2})^4\) \(=8 \sqrt{3} \sqrt{2}\left[(\sqrt{3})^2+(\sqrt{2})^2\right]\) \(=8 \sqrt{3} \sqrt{2}[3+2]\) \(=8 \sqrt{3} \sqrt{2} \times 5\) \(=40 \sqrt{6}\)
Kerala CEE-2018
Binomial Theorem and its Simple Application
119387
Let \(t_n\) denote the \(n^{\text {th }}\) term in a binomial expansion. If \(\frac{t_6}{t_5}\) in the expansion of \((a+b)^{n+4}\) and \(\frac{t_5}{t_4}\) in the expansion of \((a+b)^n\) are equal, then \(\mathbf{n}\) is equal to
1 9
2 11
3 13
4 15
5 17
Explanation:
D \(\mathrm{t}_6\) and \(\mathrm{t}_5\) in the expansion of \((\mathrm{a}+\mathrm{b})^{\mathrm{n}+4}\) is \(t_5=t_{4+1}={ }^{n+4} C_4 a^{n+4-4} \cdot b^4={ }^{n+4} C_4 a^n b^4\) \(t_6=t_{5+1}={ }^{n+4} C_5 a^{n+4-5} \cdot b^5={ }^{n+4} C_5 a^{n-1} b^5\) \(\frac{t_6}{t_5}=\frac{{ }^{n+4} C_5 \cdot a^{n-1} \cdot b^5}{{ }^{n+4} C_4 \cdot a^n \cdot b^4}=\frac{{ }^{n+4} C_5}{{ }^{n+4} C_4}\left(\frac{b}{a}\right)=\frac{n}{5}\left(\frac{b}{a}\right)\) \(\because \mathrm{t}_5\) and \(\mathrm{t}_4\) in the expansion of \((a+b)^n\) is \(t_5=t_{4+1}={ }^n C_4 \cdot a^{n-4} \cdot b^4\) \(t_4=t_{3+1}={ }^n C_3 \cdot a^{n-3} b^3\) \(\frac{t_5}{t_4}=\frac{C_4 \cdot a^{n-4} \cdot b^4}{{ }^n C_3 \cdot a^{n-3} \cdot b^3}=\frac{n-3}{4}\left(\frac{b}{a}\right)\) On equating equation (i) and (ii), we get - \(\frac{\mathrm{n}}{5}\left(-\frac{\mathrm{b}}{\mathrm{a}}\right)=\frac{\mathrm{n}-3}{4}\left(\frac{\mathrm{b}}{\mathrm{a}}\right)\) \(4 \mathrm{n}=5 \mathrm{n}-15\) \(\mathrm{n}=15\)
Kerala CEE-2013
Binomial Theorem and its Simple Application
119388
If \(\left(1+x+x^2\right)^n=1+a_1 x+a_2 x^2+\ldots .+a_{2 n} x^{2 n}\), then \(2 a_1-3 a_2+\ldots-(2 n+1) a_{2 n}\) is equal to
119389
If the average of the numbers \(1,2,3, \ldots 98,99, x\) is \(100 x\), then the value of \(x\) is
1 \(\frac{51}{100}\)
2 \(\frac{50}{99}\)
3 \(\frac{1}{2}\)
4 \(\frac{51}{99}\)
5 \(\frac{50}{101}\)
Explanation:
E Given, The average of the no. \(1,2,3, \ldots . .98,99, \mathrm{x}\) is \(100 \mathrm{x}\) \(\frac{(1+2+3+\ldots .+99)}{10}=100 \mathrm{x}\) \(\frac{99}{2}(1+99)+\mathrm{x}=10000 \mathrm{x}\) \(99 \times 50=9999 \mathrm{x}\) \(\mathrm{x}=\frac{99 \times 50}{9999}\) \(\mathrm{x}=\frac{50}{101}\)
D : Given, \((\sqrt{3}+\sqrt{2})^4-(\sqrt{3}-\sqrt{2})^4\) \((x+y)^4-(x-y)^4=8 x^3 y+8 x y^3\) \(=8 x y\left(x^2+y^2\right)\) \(x=\sqrt{3}, y=\sqrt{2}\) \((\sqrt{3}+\sqrt{2})^4-(\sqrt{3}-\sqrt{2})^4\) \(=8 \sqrt{3} \sqrt{2}\left[(\sqrt{3})^2+(\sqrt{2})^2\right]\) \(=8 \sqrt{3} \sqrt{2}[3+2]\) \(=8 \sqrt{3} \sqrt{2} \times 5\) \(=40 \sqrt{6}\)
Kerala CEE-2018
Binomial Theorem and its Simple Application
119387
Let \(t_n\) denote the \(n^{\text {th }}\) term in a binomial expansion. If \(\frac{t_6}{t_5}\) in the expansion of \((a+b)^{n+4}\) and \(\frac{t_5}{t_4}\) in the expansion of \((a+b)^n\) are equal, then \(\mathbf{n}\) is equal to
1 9
2 11
3 13
4 15
5 17
Explanation:
D \(\mathrm{t}_6\) and \(\mathrm{t}_5\) in the expansion of \((\mathrm{a}+\mathrm{b})^{\mathrm{n}+4}\) is \(t_5=t_{4+1}={ }^{n+4} C_4 a^{n+4-4} \cdot b^4={ }^{n+4} C_4 a^n b^4\) \(t_6=t_{5+1}={ }^{n+4} C_5 a^{n+4-5} \cdot b^5={ }^{n+4} C_5 a^{n-1} b^5\) \(\frac{t_6}{t_5}=\frac{{ }^{n+4} C_5 \cdot a^{n-1} \cdot b^5}{{ }^{n+4} C_4 \cdot a^n \cdot b^4}=\frac{{ }^{n+4} C_5}{{ }^{n+4} C_4}\left(\frac{b}{a}\right)=\frac{n}{5}\left(\frac{b}{a}\right)\) \(\because \mathrm{t}_5\) and \(\mathrm{t}_4\) in the expansion of \((a+b)^n\) is \(t_5=t_{4+1}={ }^n C_4 \cdot a^{n-4} \cdot b^4\) \(t_4=t_{3+1}={ }^n C_3 \cdot a^{n-3} b^3\) \(\frac{t_5}{t_4}=\frac{C_4 \cdot a^{n-4} \cdot b^4}{{ }^n C_3 \cdot a^{n-3} \cdot b^3}=\frac{n-3}{4}\left(\frac{b}{a}\right)\) On equating equation (i) and (ii), we get - \(\frac{\mathrm{n}}{5}\left(-\frac{\mathrm{b}}{\mathrm{a}}\right)=\frac{\mathrm{n}-3}{4}\left(\frac{\mathrm{b}}{\mathrm{a}}\right)\) \(4 \mathrm{n}=5 \mathrm{n}-15\) \(\mathrm{n}=15\)
Kerala CEE-2013
Binomial Theorem and its Simple Application
119388
If \(\left(1+x+x^2\right)^n=1+a_1 x+a_2 x^2+\ldots .+a_{2 n} x^{2 n}\), then \(2 a_1-3 a_2+\ldots-(2 n+1) a_{2 n}\) is equal to
119389
If the average of the numbers \(1,2,3, \ldots 98,99, x\) is \(100 x\), then the value of \(x\) is
1 \(\frac{51}{100}\)
2 \(\frac{50}{99}\)
3 \(\frac{1}{2}\)
4 \(\frac{51}{99}\)
5 \(\frac{50}{101}\)
Explanation:
E Given, The average of the no. \(1,2,3, \ldots . .98,99, \mathrm{x}\) is \(100 \mathrm{x}\) \(\frac{(1+2+3+\ldots .+99)}{10}=100 \mathrm{x}\) \(\frac{99}{2}(1+99)+\mathrm{x}=10000 \mathrm{x}\) \(99 \times 50=9999 \mathrm{x}\) \(\mathrm{x}=\frac{99 \times 50}{9999}\) \(\mathrm{x}=\frac{50}{101}\)
D : Given, \((\sqrt{3}+\sqrt{2})^4-(\sqrt{3}-\sqrt{2})^4\) \((x+y)^4-(x-y)^4=8 x^3 y+8 x y^3\) \(=8 x y\left(x^2+y^2\right)\) \(x=\sqrt{3}, y=\sqrt{2}\) \((\sqrt{3}+\sqrt{2})^4-(\sqrt{3}-\sqrt{2})^4\) \(=8 \sqrt{3} \sqrt{2}\left[(\sqrt{3})^2+(\sqrt{2})^2\right]\) \(=8 \sqrt{3} \sqrt{2}[3+2]\) \(=8 \sqrt{3} \sqrt{2} \times 5\) \(=40 \sqrt{6}\)
Kerala CEE-2018
Binomial Theorem and its Simple Application
119387
Let \(t_n\) denote the \(n^{\text {th }}\) term in a binomial expansion. If \(\frac{t_6}{t_5}\) in the expansion of \((a+b)^{n+4}\) and \(\frac{t_5}{t_4}\) in the expansion of \((a+b)^n\) are equal, then \(\mathbf{n}\) is equal to
1 9
2 11
3 13
4 15
5 17
Explanation:
D \(\mathrm{t}_6\) and \(\mathrm{t}_5\) in the expansion of \((\mathrm{a}+\mathrm{b})^{\mathrm{n}+4}\) is \(t_5=t_{4+1}={ }^{n+4} C_4 a^{n+4-4} \cdot b^4={ }^{n+4} C_4 a^n b^4\) \(t_6=t_{5+1}={ }^{n+4} C_5 a^{n+4-5} \cdot b^5={ }^{n+4} C_5 a^{n-1} b^5\) \(\frac{t_6}{t_5}=\frac{{ }^{n+4} C_5 \cdot a^{n-1} \cdot b^5}{{ }^{n+4} C_4 \cdot a^n \cdot b^4}=\frac{{ }^{n+4} C_5}{{ }^{n+4} C_4}\left(\frac{b}{a}\right)=\frac{n}{5}\left(\frac{b}{a}\right)\) \(\because \mathrm{t}_5\) and \(\mathrm{t}_4\) in the expansion of \((a+b)^n\) is \(t_5=t_{4+1}={ }^n C_4 \cdot a^{n-4} \cdot b^4\) \(t_4=t_{3+1}={ }^n C_3 \cdot a^{n-3} b^3\) \(\frac{t_5}{t_4}=\frac{C_4 \cdot a^{n-4} \cdot b^4}{{ }^n C_3 \cdot a^{n-3} \cdot b^3}=\frac{n-3}{4}\left(\frac{b}{a}\right)\) On equating equation (i) and (ii), we get - \(\frac{\mathrm{n}}{5}\left(-\frac{\mathrm{b}}{\mathrm{a}}\right)=\frac{\mathrm{n}-3}{4}\left(\frac{\mathrm{b}}{\mathrm{a}}\right)\) \(4 \mathrm{n}=5 \mathrm{n}-15\) \(\mathrm{n}=15\)
Kerala CEE-2013
Binomial Theorem and its Simple Application
119388
If \(\left(1+x+x^2\right)^n=1+a_1 x+a_2 x^2+\ldots .+a_{2 n} x^{2 n}\), then \(2 a_1-3 a_2+\ldots-(2 n+1) a_{2 n}\) is equal to
119389
If the average of the numbers \(1,2,3, \ldots 98,99, x\) is \(100 x\), then the value of \(x\) is
1 \(\frac{51}{100}\)
2 \(\frac{50}{99}\)
3 \(\frac{1}{2}\)
4 \(\frac{51}{99}\)
5 \(\frac{50}{101}\)
Explanation:
E Given, The average of the no. \(1,2,3, \ldots . .98,99, \mathrm{x}\) is \(100 \mathrm{x}\) \(\frac{(1+2+3+\ldots .+99)}{10}=100 \mathrm{x}\) \(\frac{99}{2}(1+99)+\mathrm{x}=10000 \mathrm{x}\) \(99 \times 50=9999 \mathrm{x}\) \(\mathrm{x}=\frac{99 \times 50}{9999}\) \(\mathrm{x}=\frac{50}{101}\)
D : Given, \((\sqrt{3}+\sqrt{2})^4-(\sqrt{3}-\sqrt{2})^4\) \((x+y)^4-(x-y)^4=8 x^3 y+8 x y^3\) \(=8 x y\left(x^2+y^2\right)\) \(x=\sqrt{3}, y=\sqrt{2}\) \((\sqrt{3}+\sqrt{2})^4-(\sqrt{3}-\sqrt{2})^4\) \(=8 \sqrt{3} \sqrt{2}\left[(\sqrt{3})^2+(\sqrt{2})^2\right]\) \(=8 \sqrt{3} \sqrt{2}[3+2]\) \(=8 \sqrt{3} \sqrt{2} \times 5\) \(=40 \sqrt{6}\)
Kerala CEE-2018
Binomial Theorem and its Simple Application
119387
Let \(t_n\) denote the \(n^{\text {th }}\) term in a binomial expansion. If \(\frac{t_6}{t_5}\) in the expansion of \((a+b)^{n+4}\) and \(\frac{t_5}{t_4}\) in the expansion of \((a+b)^n\) are equal, then \(\mathbf{n}\) is equal to
1 9
2 11
3 13
4 15
5 17
Explanation:
D \(\mathrm{t}_6\) and \(\mathrm{t}_5\) in the expansion of \((\mathrm{a}+\mathrm{b})^{\mathrm{n}+4}\) is \(t_5=t_{4+1}={ }^{n+4} C_4 a^{n+4-4} \cdot b^4={ }^{n+4} C_4 a^n b^4\) \(t_6=t_{5+1}={ }^{n+4} C_5 a^{n+4-5} \cdot b^5={ }^{n+4} C_5 a^{n-1} b^5\) \(\frac{t_6}{t_5}=\frac{{ }^{n+4} C_5 \cdot a^{n-1} \cdot b^5}{{ }^{n+4} C_4 \cdot a^n \cdot b^4}=\frac{{ }^{n+4} C_5}{{ }^{n+4} C_4}\left(\frac{b}{a}\right)=\frac{n}{5}\left(\frac{b}{a}\right)\) \(\because \mathrm{t}_5\) and \(\mathrm{t}_4\) in the expansion of \((a+b)^n\) is \(t_5=t_{4+1}={ }^n C_4 \cdot a^{n-4} \cdot b^4\) \(t_4=t_{3+1}={ }^n C_3 \cdot a^{n-3} b^3\) \(\frac{t_5}{t_4}=\frac{C_4 \cdot a^{n-4} \cdot b^4}{{ }^n C_3 \cdot a^{n-3} \cdot b^3}=\frac{n-3}{4}\left(\frac{b}{a}\right)\) On equating equation (i) and (ii), we get - \(\frac{\mathrm{n}}{5}\left(-\frac{\mathrm{b}}{\mathrm{a}}\right)=\frac{\mathrm{n}-3}{4}\left(\frac{\mathrm{b}}{\mathrm{a}}\right)\) \(4 \mathrm{n}=5 \mathrm{n}-15\) \(\mathrm{n}=15\)
Kerala CEE-2013
Binomial Theorem and its Simple Application
119388
If \(\left(1+x+x^2\right)^n=1+a_1 x+a_2 x^2+\ldots .+a_{2 n} x^{2 n}\), then \(2 a_1-3 a_2+\ldots-(2 n+1) a_{2 n}\) is equal to
119389
If the average of the numbers \(1,2,3, \ldots 98,99, x\) is \(100 x\), then the value of \(x\) is
1 \(\frac{51}{100}\)
2 \(\frac{50}{99}\)
3 \(\frac{1}{2}\)
4 \(\frac{51}{99}\)
5 \(\frac{50}{101}\)
Explanation:
E Given, The average of the no. \(1,2,3, \ldots . .98,99, \mathrm{x}\) is \(100 \mathrm{x}\) \(\frac{(1+2+3+\ldots .+99)}{10}=100 \mathrm{x}\) \(\frac{99}{2}(1+99)+\mathrm{x}=10000 \mathrm{x}\) \(99 \times 50=9999 \mathrm{x}\) \(\mathrm{x}=\frac{99 \times 50}{9999}\) \(\mathrm{x}=\frac{50}{101}\)