119341
The ninth term in the expansion of \(\left[3^{\log _3 \sqrt{25^{x-1}+7}}+3^{\left(\frac{-1}{8}\right) \log _3\left(5^{x-1}+1\right)}\right]^{10} \text { is equal to } 180 \text {, then }\) \(x\) is equal to
1 1
2 2
3 3
4 None of these
Explanation:
A Given, \(\left[3^{\log _3 \sqrt{25^{x-1}+7}}+3^{\left(\frac{-1}{8}\right)^{\log _3\left(5^{x-1}+1\right)}}\right]^{10}\) \(=\left[\sqrt{25^{x-1}+7}+\left(5^{x-1}+1\right)^{\frac{-1}{8}}\right]^{10}\) Since, \(\mathrm{T}_9=\mathrm{T}_{8+1}\) term \(=180\) \(\mathrm{T}_{\mathrm{r}+1}=\mathrm{n}_{\mathrm{C}_{\mathrm{r}}} \mathrm{x}^{\mathrm{n}-\mathrm{r}} \mathrm{a}^{\mathrm{r}}\) So, \(\mathrm{T}_{8+1}={ }^{10} \mathrm{C}_8\left(\sqrt{25^{\mathrm{x}-1}+7}\right)^2 \times\left(5^{\mathrm{x}-1}+1\right)^{\left(\frac{-1 \times 8}{8}\right)}=180\) \({ }^{10} \mathrm{C}_8 \times\left(\frac{25^{\mathrm{x}-1}+7}{5^{\mathrm{x}-1}+1}\right)=180\) \(45 \times \frac{25^{\mathrm{x}-1}+7}{5^{\mathrm{x}-1}+1}=180\) Let, put \(5^{\mathrm{x}-1}=\mathrm{z}\) Then, \(\frac{z^2+7}{z+1}=4\) \(z^2-4 z+3=0\) \(z^2-3 z-z+3=0\) \((z-1)(z-3)=0\) \(z=1 \text { or } z=3\) So, \(5^{x-1}=(5)^0\) \(x-1=0\) \(x=1\)
BCECE-2010
Binomial Theorem and its Simple Application
119342
If the expansion in powers of \(x\) of the function \(\frac{1}{(1-a x)(1-b x)}\) is \(a_0+a_1 x+a_2 x^2+a_3 x^3+\ldots . .\), where \(|a x|,|b x|\lt 1\), then what is \(a_n\) equal to?
C The given function, \(\frac{1}{(1-a x)(1-b x)}\) \(\Rightarrow(1-a x)^{-1}(1-b x)^{-1}\) By the binomial expansion \(\left(1+a x+a^2 x^2+a^3 x^3 \ldots\right)\left(1+b x+b^2 x^2 \ldots \ldots\right)\) Hence, \(a_n=\) coefficient of \(x^n\) in the expansion \((1-a x)^{-1}(1-b x)^{-1}\) \(=a^0 b^n+a b^{n-1}+\ldots .+a^n b^0\) The given term common ratio b/a \(a_n=a_n \frac{\left[1-(b / a)^{n+1}\right]}{1-b / a} =\frac{a^n\left(a^{n+1}-b^{n+1}\right) / a^{n+1}}{\frac{(a-b)}{a}}\) \(=\frac{a^{n+1}-b^{n+1}}{a-b}\) \(=\frac{b^{n+1}-a^{n+1}}{b-a}\)
SCRA-2009]
Binomial Theorem and its Simple Application
119343
\(\mathrm{C}_1+2 \mathrm{C}_2+3 \mathrm{C}_3+\ldots+\mathrm{nC}_{\mathrm{n}}\) is equal to
119344
Coefficient of \(x^n\) in the expansion of \(1+\frac{\mathbf{a}+\mathbf{b x}}{1 !}+\frac{(\mathbf{a}+\mathbf{b x})^2}{2 !}+\frac{(\mathbf{a}+\mathbf{b x})^3}{3 !}+\ldots\) is
119341
The ninth term in the expansion of \(\left[3^{\log _3 \sqrt{25^{x-1}+7}}+3^{\left(\frac{-1}{8}\right) \log _3\left(5^{x-1}+1\right)}\right]^{10} \text { is equal to } 180 \text {, then }\) \(x\) is equal to
1 1
2 2
3 3
4 None of these
Explanation:
A Given, \(\left[3^{\log _3 \sqrt{25^{x-1}+7}}+3^{\left(\frac{-1}{8}\right)^{\log _3\left(5^{x-1}+1\right)}}\right]^{10}\) \(=\left[\sqrt{25^{x-1}+7}+\left(5^{x-1}+1\right)^{\frac{-1}{8}}\right]^{10}\) Since, \(\mathrm{T}_9=\mathrm{T}_{8+1}\) term \(=180\) \(\mathrm{T}_{\mathrm{r}+1}=\mathrm{n}_{\mathrm{C}_{\mathrm{r}}} \mathrm{x}^{\mathrm{n}-\mathrm{r}} \mathrm{a}^{\mathrm{r}}\) So, \(\mathrm{T}_{8+1}={ }^{10} \mathrm{C}_8\left(\sqrt{25^{\mathrm{x}-1}+7}\right)^2 \times\left(5^{\mathrm{x}-1}+1\right)^{\left(\frac{-1 \times 8}{8}\right)}=180\) \({ }^{10} \mathrm{C}_8 \times\left(\frac{25^{\mathrm{x}-1}+7}{5^{\mathrm{x}-1}+1}\right)=180\) \(45 \times \frac{25^{\mathrm{x}-1}+7}{5^{\mathrm{x}-1}+1}=180\) Let, put \(5^{\mathrm{x}-1}=\mathrm{z}\) Then, \(\frac{z^2+7}{z+1}=4\) \(z^2-4 z+3=0\) \(z^2-3 z-z+3=0\) \((z-1)(z-3)=0\) \(z=1 \text { or } z=3\) So, \(5^{x-1}=(5)^0\) \(x-1=0\) \(x=1\)
BCECE-2010
Binomial Theorem and its Simple Application
119342
If the expansion in powers of \(x\) of the function \(\frac{1}{(1-a x)(1-b x)}\) is \(a_0+a_1 x+a_2 x^2+a_3 x^3+\ldots . .\), where \(|a x|,|b x|\lt 1\), then what is \(a_n\) equal to?
C The given function, \(\frac{1}{(1-a x)(1-b x)}\) \(\Rightarrow(1-a x)^{-1}(1-b x)^{-1}\) By the binomial expansion \(\left(1+a x+a^2 x^2+a^3 x^3 \ldots\right)\left(1+b x+b^2 x^2 \ldots \ldots\right)\) Hence, \(a_n=\) coefficient of \(x^n\) in the expansion \((1-a x)^{-1}(1-b x)^{-1}\) \(=a^0 b^n+a b^{n-1}+\ldots .+a^n b^0\) The given term common ratio b/a \(a_n=a_n \frac{\left[1-(b / a)^{n+1}\right]}{1-b / a} =\frac{a^n\left(a^{n+1}-b^{n+1}\right) / a^{n+1}}{\frac{(a-b)}{a}}\) \(=\frac{a^{n+1}-b^{n+1}}{a-b}\) \(=\frac{b^{n+1}-a^{n+1}}{b-a}\)
SCRA-2009]
Binomial Theorem and its Simple Application
119343
\(\mathrm{C}_1+2 \mathrm{C}_2+3 \mathrm{C}_3+\ldots+\mathrm{nC}_{\mathrm{n}}\) is equal to
119344
Coefficient of \(x^n\) in the expansion of \(1+\frac{\mathbf{a}+\mathbf{b x}}{1 !}+\frac{(\mathbf{a}+\mathbf{b x})^2}{2 !}+\frac{(\mathbf{a}+\mathbf{b x})^3}{3 !}+\ldots\) is
NEET Test Series from KOTA - 10 Papers In MS WORD
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Binomial Theorem and its Simple Application
119341
The ninth term in the expansion of \(\left[3^{\log _3 \sqrt{25^{x-1}+7}}+3^{\left(\frac{-1}{8}\right) \log _3\left(5^{x-1}+1\right)}\right]^{10} \text { is equal to } 180 \text {, then }\) \(x\) is equal to
1 1
2 2
3 3
4 None of these
Explanation:
A Given, \(\left[3^{\log _3 \sqrt{25^{x-1}+7}}+3^{\left(\frac{-1}{8}\right)^{\log _3\left(5^{x-1}+1\right)}}\right]^{10}\) \(=\left[\sqrt{25^{x-1}+7}+\left(5^{x-1}+1\right)^{\frac{-1}{8}}\right]^{10}\) Since, \(\mathrm{T}_9=\mathrm{T}_{8+1}\) term \(=180\) \(\mathrm{T}_{\mathrm{r}+1}=\mathrm{n}_{\mathrm{C}_{\mathrm{r}}} \mathrm{x}^{\mathrm{n}-\mathrm{r}} \mathrm{a}^{\mathrm{r}}\) So, \(\mathrm{T}_{8+1}={ }^{10} \mathrm{C}_8\left(\sqrt{25^{\mathrm{x}-1}+7}\right)^2 \times\left(5^{\mathrm{x}-1}+1\right)^{\left(\frac{-1 \times 8}{8}\right)}=180\) \({ }^{10} \mathrm{C}_8 \times\left(\frac{25^{\mathrm{x}-1}+7}{5^{\mathrm{x}-1}+1}\right)=180\) \(45 \times \frac{25^{\mathrm{x}-1}+7}{5^{\mathrm{x}-1}+1}=180\) Let, put \(5^{\mathrm{x}-1}=\mathrm{z}\) Then, \(\frac{z^2+7}{z+1}=4\) \(z^2-4 z+3=0\) \(z^2-3 z-z+3=0\) \((z-1)(z-3)=0\) \(z=1 \text { or } z=3\) So, \(5^{x-1}=(5)^0\) \(x-1=0\) \(x=1\)
BCECE-2010
Binomial Theorem and its Simple Application
119342
If the expansion in powers of \(x\) of the function \(\frac{1}{(1-a x)(1-b x)}\) is \(a_0+a_1 x+a_2 x^2+a_3 x^3+\ldots . .\), where \(|a x|,|b x|\lt 1\), then what is \(a_n\) equal to?
C The given function, \(\frac{1}{(1-a x)(1-b x)}\) \(\Rightarrow(1-a x)^{-1}(1-b x)^{-1}\) By the binomial expansion \(\left(1+a x+a^2 x^2+a^3 x^3 \ldots\right)\left(1+b x+b^2 x^2 \ldots \ldots\right)\) Hence, \(a_n=\) coefficient of \(x^n\) in the expansion \((1-a x)^{-1}(1-b x)^{-1}\) \(=a^0 b^n+a b^{n-1}+\ldots .+a^n b^0\) The given term common ratio b/a \(a_n=a_n \frac{\left[1-(b / a)^{n+1}\right]}{1-b / a} =\frac{a^n\left(a^{n+1}-b^{n+1}\right) / a^{n+1}}{\frac{(a-b)}{a}}\) \(=\frac{a^{n+1}-b^{n+1}}{a-b}\) \(=\frac{b^{n+1}-a^{n+1}}{b-a}\)
SCRA-2009]
Binomial Theorem and its Simple Application
119343
\(\mathrm{C}_1+2 \mathrm{C}_2+3 \mathrm{C}_3+\ldots+\mathrm{nC}_{\mathrm{n}}\) is equal to
119344
Coefficient of \(x^n\) in the expansion of \(1+\frac{\mathbf{a}+\mathbf{b x}}{1 !}+\frac{(\mathbf{a}+\mathbf{b x})^2}{2 !}+\frac{(\mathbf{a}+\mathbf{b x})^3}{3 !}+\ldots\) is
119341
The ninth term in the expansion of \(\left[3^{\log _3 \sqrt{25^{x-1}+7}}+3^{\left(\frac{-1}{8}\right) \log _3\left(5^{x-1}+1\right)}\right]^{10} \text { is equal to } 180 \text {, then }\) \(x\) is equal to
1 1
2 2
3 3
4 None of these
Explanation:
A Given, \(\left[3^{\log _3 \sqrt{25^{x-1}+7}}+3^{\left(\frac{-1}{8}\right)^{\log _3\left(5^{x-1}+1\right)}}\right]^{10}\) \(=\left[\sqrt{25^{x-1}+7}+\left(5^{x-1}+1\right)^{\frac{-1}{8}}\right]^{10}\) Since, \(\mathrm{T}_9=\mathrm{T}_{8+1}\) term \(=180\) \(\mathrm{T}_{\mathrm{r}+1}=\mathrm{n}_{\mathrm{C}_{\mathrm{r}}} \mathrm{x}^{\mathrm{n}-\mathrm{r}} \mathrm{a}^{\mathrm{r}}\) So, \(\mathrm{T}_{8+1}={ }^{10} \mathrm{C}_8\left(\sqrt{25^{\mathrm{x}-1}+7}\right)^2 \times\left(5^{\mathrm{x}-1}+1\right)^{\left(\frac{-1 \times 8}{8}\right)}=180\) \({ }^{10} \mathrm{C}_8 \times\left(\frac{25^{\mathrm{x}-1}+7}{5^{\mathrm{x}-1}+1}\right)=180\) \(45 \times \frac{25^{\mathrm{x}-1}+7}{5^{\mathrm{x}-1}+1}=180\) Let, put \(5^{\mathrm{x}-1}=\mathrm{z}\) Then, \(\frac{z^2+7}{z+1}=4\) \(z^2-4 z+3=0\) \(z^2-3 z-z+3=0\) \((z-1)(z-3)=0\) \(z=1 \text { or } z=3\) So, \(5^{x-1}=(5)^0\) \(x-1=0\) \(x=1\)
BCECE-2010
Binomial Theorem and its Simple Application
119342
If the expansion in powers of \(x\) of the function \(\frac{1}{(1-a x)(1-b x)}\) is \(a_0+a_1 x+a_2 x^2+a_3 x^3+\ldots . .\), where \(|a x|,|b x|\lt 1\), then what is \(a_n\) equal to?
C The given function, \(\frac{1}{(1-a x)(1-b x)}\) \(\Rightarrow(1-a x)^{-1}(1-b x)^{-1}\) By the binomial expansion \(\left(1+a x+a^2 x^2+a^3 x^3 \ldots\right)\left(1+b x+b^2 x^2 \ldots \ldots\right)\) Hence, \(a_n=\) coefficient of \(x^n\) in the expansion \((1-a x)^{-1}(1-b x)^{-1}\) \(=a^0 b^n+a b^{n-1}+\ldots .+a^n b^0\) The given term common ratio b/a \(a_n=a_n \frac{\left[1-(b / a)^{n+1}\right]}{1-b / a} =\frac{a^n\left(a^{n+1}-b^{n+1}\right) / a^{n+1}}{\frac{(a-b)}{a}}\) \(=\frac{a^{n+1}-b^{n+1}}{a-b}\) \(=\frac{b^{n+1}-a^{n+1}}{b-a}\)
SCRA-2009]
Binomial Theorem and its Simple Application
119343
\(\mathrm{C}_1+2 \mathrm{C}_2+3 \mathrm{C}_3+\ldots+\mathrm{nC}_{\mathrm{n}}\) is equal to
119344
Coefficient of \(x^n\) in the expansion of \(1+\frac{\mathbf{a}+\mathbf{b x}}{1 !}+\frac{(\mathbf{a}+\mathbf{b x})^2}{2 !}+\frac{(\mathbf{a}+\mathbf{b x})^3}{3 !}+\ldots\) is