119354
If \(\frac{(1-p x)^{-1}}{(1-q x)}=a_0+a_1 x+a_2 x^2+a_3 x^3+\ldots\). then \(a_n=\)
1 \(\frac{p^{n+1}-q^{n+1}}{q-p}\)
2 \(\frac{p^{n+1}-q^{n+1}}{p-q}\)
3 \(\frac{p^n-q^n}{q-p}\)
4 \(\frac{p^n-q^n}{p-q}\)
Explanation:
B Given,\(\frac{(1-p x)^{-1}}{(1-q x)}=a_0+a_1 x+a_2 x^2+a_3 x^3+\ldots .\) \(\because(1-\mathrm{px})^{-1}=1+\mathrm{px}+\mathrm{p}^2 \mathrm{x}^2+\mathrm{p}^3 \mathrm{x}^3+\ldots \ldots+\mathrm{p}^{\mathrm{n}} \mathrm{x}^{\mathrm{n}} \ldots \ldots\) \(\text { and }(1-q x)^{-1}=1+q x+q^2 x^2+q^3 x^3+\ldots+q^n x^n \ldots\) \((1-p x)^{-1}(1-q x)^{-1}=p^n+p^{n-1} \cdot q+p^{n-2} \cdot q^2+p^{n-}\) \({ }^3 \cdot \mathrm{q}^3+\ldots . .+\mathrm{q}^{\mathrm{n}}\) \(a^n=\frac{P^n\left[1-\left(\frac{q}{p}\right)^{n+1}\right]}{1-q / p}\) \(a^n=\frac{p^n\left(p^{n+1}-q^{n+1}\right) p}{(p-q) p^{n+1}}\) \(=\frac{p^{n+1}-q^{n+1}}{(p-q)}\) Then, coefficient of \(x^{\mathrm{n}}\) in the expansion, So, \(a_n=\frac{p^{n+1}-q^{n+1}}{p-q}\)
AP EAMCET-20.04.2019
Binomial Theorem and its Simple Application
119355
If \(x\) small, so the \(x^2\) and higher powers can be neglected, then the approximate value for \(\frac{(1-2 x)^{-1}(1-3 x)^{-2}}{(1-4 x)^{-3}}\) is
119356
If \((a+b x)^{-3}=\frac{1}{27}+\frac{1}{3} x+\ldots \ldots\), then the ordered pair (a, b) equals to
1 \((3,-27)\)
2 \(\left(1, \frac{1}{3}\right)\)
3 \((3,9)\)
4 \((3,-9)\)
Explanation:
D Given, expansion \((\mathrm{a}+\mathrm{bx})^{-3}=\frac{1}{27}+\left(\frac{1}{3}\right) \mathrm{x}\) Using the binomial expansion, \((1+x)^{-n}=1-n x+\frac{(n+1) n}{2 !} x^2-\frac{n(n+1(n+2))}{3 !} x^3\) \((a+b x)^{-3}=a^{-3}\left[1-3 \frac{b}{a} x\right]+6\left(\frac{b x}{a}\right)^2+\ldots . \ldots . \text { (ii) }\) On comparing the coefficient of equation (i) and (ii), we get - \(\frac{1}{a^3}=\frac{1}{27}, \frac{-3 b}{a^4}=\frac{1}{3}\) \(a=3,-3 b=\frac{3^4}{3}\) \(a=3, b=-9\)
AP EAMCET-2014
Binomial Theorem and its Simple Application
119357
The coefficient of \(x^n\) in the expansion of \(\frac{e^{7 x}+e^x}{e^{3 x}}\) is
119354
If \(\frac{(1-p x)^{-1}}{(1-q x)}=a_0+a_1 x+a_2 x^2+a_3 x^3+\ldots\). then \(a_n=\)
1 \(\frac{p^{n+1}-q^{n+1}}{q-p}\)
2 \(\frac{p^{n+1}-q^{n+1}}{p-q}\)
3 \(\frac{p^n-q^n}{q-p}\)
4 \(\frac{p^n-q^n}{p-q}\)
Explanation:
B Given,\(\frac{(1-p x)^{-1}}{(1-q x)}=a_0+a_1 x+a_2 x^2+a_3 x^3+\ldots .\) \(\because(1-\mathrm{px})^{-1}=1+\mathrm{px}+\mathrm{p}^2 \mathrm{x}^2+\mathrm{p}^3 \mathrm{x}^3+\ldots \ldots+\mathrm{p}^{\mathrm{n}} \mathrm{x}^{\mathrm{n}} \ldots \ldots\) \(\text { and }(1-q x)^{-1}=1+q x+q^2 x^2+q^3 x^3+\ldots+q^n x^n \ldots\) \((1-p x)^{-1}(1-q x)^{-1}=p^n+p^{n-1} \cdot q+p^{n-2} \cdot q^2+p^{n-}\) \({ }^3 \cdot \mathrm{q}^3+\ldots . .+\mathrm{q}^{\mathrm{n}}\) \(a^n=\frac{P^n\left[1-\left(\frac{q}{p}\right)^{n+1}\right]}{1-q / p}\) \(a^n=\frac{p^n\left(p^{n+1}-q^{n+1}\right) p}{(p-q) p^{n+1}}\) \(=\frac{p^{n+1}-q^{n+1}}{(p-q)}\) Then, coefficient of \(x^{\mathrm{n}}\) in the expansion, So, \(a_n=\frac{p^{n+1}-q^{n+1}}{p-q}\)
AP EAMCET-20.04.2019
Binomial Theorem and its Simple Application
119355
If \(x\) small, so the \(x^2\) and higher powers can be neglected, then the approximate value for \(\frac{(1-2 x)^{-1}(1-3 x)^{-2}}{(1-4 x)^{-3}}\) is
119356
If \((a+b x)^{-3}=\frac{1}{27}+\frac{1}{3} x+\ldots \ldots\), then the ordered pair (a, b) equals to
1 \((3,-27)\)
2 \(\left(1, \frac{1}{3}\right)\)
3 \((3,9)\)
4 \((3,-9)\)
Explanation:
D Given, expansion \((\mathrm{a}+\mathrm{bx})^{-3}=\frac{1}{27}+\left(\frac{1}{3}\right) \mathrm{x}\) Using the binomial expansion, \((1+x)^{-n}=1-n x+\frac{(n+1) n}{2 !} x^2-\frac{n(n+1(n+2))}{3 !} x^3\) \((a+b x)^{-3}=a^{-3}\left[1-3 \frac{b}{a} x\right]+6\left(\frac{b x}{a}\right)^2+\ldots . \ldots . \text { (ii) }\) On comparing the coefficient of equation (i) and (ii), we get - \(\frac{1}{a^3}=\frac{1}{27}, \frac{-3 b}{a^4}=\frac{1}{3}\) \(a=3,-3 b=\frac{3^4}{3}\) \(a=3, b=-9\)
AP EAMCET-2014
Binomial Theorem and its Simple Application
119357
The coefficient of \(x^n\) in the expansion of \(\frac{e^{7 x}+e^x}{e^{3 x}}\) is
119354
If \(\frac{(1-p x)^{-1}}{(1-q x)}=a_0+a_1 x+a_2 x^2+a_3 x^3+\ldots\). then \(a_n=\)
1 \(\frac{p^{n+1}-q^{n+1}}{q-p}\)
2 \(\frac{p^{n+1}-q^{n+1}}{p-q}\)
3 \(\frac{p^n-q^n}{q-p}\)
4 \(\frac{p^n-q^n}{p-q}\)
Explanation:
B Given,\(\frac{(1-p x)^{-1}}{(1-q x)}=a_0+a_1 x+a_2 x^2+a_3 x^3+\ldots .\) \(\because(1-\mathrm{px})^{-1}=1+\mathrm{px}+\mathrm{p}^2 \mathrm{x}^2+\mathrm{p}^3 \mathrm{x}^3+\ldots \ldots+\mathrm{p}^{\mathrm{n}} \mathrm{x}^{\mathrm{n}} \ldots \ldots\) \(\text { and }(1-q x)^{-1}=1+q x+q^2 x^2+q^3 x^3+\ldots+q^n x^n \ldots\) \((1-p x)^{-1}(1-q x)^{-1}=p^n+p^{n-1} \cdot q+p^{n-2} \cdot q^2+p^{n-}\) \({ }^3 \cdot \mathrm{q}^3+\ldots . .+\mathrm{q}^{\mathrm{n}}\) \(a^n=\frac{P^n\left[1-\left(\frac{q}{p}\right)^{n+1}\right]}{1-q / p}\) \(a^n=\frac{p^n\left(p^{n+1}-q^{n+1}\right) p}{(p-q) p^{n+1}}\) \(=\frac{p^{n+1}-q^{n+1}}{(p-q)}\) Then, coefficient of \(x^{\mathrm{n}}\) in the expansion, So, \(a_n=\frac{p^{n+1}-q^{n+1}}{p-q}\)
AP EAMCET-20.04.2019
Binomial Theorem and its Simple Application
119355
If \(x\) small, so the \(x^2\) and higher powers can be neglected, then the approximate value for \(\frac{(1-2 x)^{-1}(1-3 x)^{-2}}{(1-4 x)^{-3}}\) is
119356
If \((a+b x)^{-3}=\frac{1}{27}+\frac{1}{3} x+\ldots \ldots\), then the ordered pair (a, b) equals to
1 \((3,-27)\)
2 \(\left(1, \frac{1}{3}\right)\)
3 \((3,9)\)
4 \((3,-9)\)
Explanation:
D Given, expansion \((\mathrm{a}+\mathrm{bx})^{-3}=\frac{1}{27}+\left(\frac{1}{3}\right) \mathrm{x}\) Using the binomial expansion, \((1+x)^{-n}=1-n x+\frac{(n+1) n}{2 !} x^2-\frac{n(n+1(n+2))}{3 !} x^3\) \((a+b x)^{-3}=a^{-3}\left[1-3 \frac{b}{a} x\right]+6\left(\frac{b x}{a}\right)^2+\ldots . \ldots . \text { (ii) }\) On comparing the coefficient of equation (i) and (ii), we get - \(\frac{1}{a^3}=\frac{1}{27}, \frac{-3 b}{a^4}=\frac{1}{3}\) \(a=3,-3 b=\frac{3^4}{3}\) \(a=3, b=-9\)
AP EAMCET-2014
Binomial Theorem and its Simple Application
119357
The coefficient of \(x^n\) in the expansion of \(\frac{e^{7 x}+e^x}{e^{3 x}}\) is
119354
If \(\frac{(1-p x)^{-1}}{(1-q x)}=a_0+a_1 x+a_2 x^2+a_3 x^3+\ldots\). then \(a_n=\)
1 \(\frac{p^{n+1}-q^{n+1}}{q-p}\)
2 \(\frac{p^{n+1}-q^{n+1}}{p-q}\)
3 \(\frac{p^n-q^n}{q-p}\)
4 \(\frac{p^n-q^n}{p-q}\)
Explanation:
B Given,\(\frac{(1-p x)^{-1}}{(1-q x)}=a_0+a_1 x+a_2 x^2+a_3 x^3+\ldots .\) \(\because(1-\mathrm{px})^{-1}=1+\mathrm{px}+\mathrm{p}^2 \mathrm{x}^2+\mathrm{p}^3 \mathrm{x}^3+\ldots \ldots+\mathrm{p}^{\mathrm{n}} \mathrm{x}^{\mathrm{n}} \ldots \ldots\) \(\text { and }(1-q x)^{-1}=1+q x+q^2 x^2+q^3 x^3+\ldots+q^n x^n \ldots\) \((1-p x)^{-1}(1-q x)^{-1}=p^n+p^{n-1} \cdot q+p^{n-2} \cdot q^2+p^{n-}\) \({ }^3 \cdot \mathrm{q}^3+\ldots . .+\mathrm{q}^{\mathrm{n}}\) \(a^n=\frac{P^n\left[1-\left(\frac{q}{p}\right)^{n+1}\right]}{1-q / p}\) \(a^n=\frac{p^n\left(p^{n+1}-q^{n+1}\right) p}{(p-q) p^{n+1}}\) \(=\frac{p^{n+1}-q^{n+1}}{(p-q)}\) Then, coefficient of \(x^{\mathrm{n}}\) in the expansion, So, \(a_n=\frac{p^{n+1}-q^{n+1}}{p-q}\)
AP EAMCET-20.04.2019
Binomial Theorem and its Simple Application
119355
If \(x\) small, so the \(x^2\) and higher powers can be neglected, then the approximate value for \(\frac{(1-2 x)^{-1}(1-3 x)^{-2}}{(1-4 x)^{-3}}\) is
119356
If \((a+b x)^{-3}=\frac{1}{27}+\frac{1}{3} x+\ldots \ldots\), then the ordered pair (a, b) equals to
1 \((3,-27)\)
2 \(\left(1, \frac{1}{3}\right)\)
3 \((3,9)\)
4 \((3,-9)\)
Explanation:
D Given, expansion \((\mathrm{a}+\mathrm{bx})^{-3}=\frac{1}{27}+\left(\frac{1}{3}\right) \mathrm{x}\) Using the binomial expansion, \((1+x)^{-n}=1-n x+\frac{(n+1) n}{2 !} x^2-\frac{n(n+1(n+2))}{3 !} x^3\) \((a+b x)^{-3}=a^{-3}\left[1-3 \frac{b}{a} x\right]+6\left(\frac{b x}{a}\right)^2+\ldots . \ldots . \text { (ii) }\) On comparing the coefficient of equation (i) and (ii), we get - \(\frac{1}{a^3}=\frac{1}{27}, \frac{-3 b}{a^4}=\frac{1}{3}\) \(a=3,-3 b=\frac{3^4}{3}\) \(a=3, b=-9\)
AP EAMCET-2014
Binomial Theorem and its Simple Application
119357
The coefficient of \(x^n\) in the expansion of \(\frac{e^{7 x}+e^x}{e^{3 x}}\) is
