JEE Main & Advanced Mathematics Sequence & Series Question Bank

done Critical Thinking Total Questions - 20

• question_answer1)
  \[\frac{\frac{1}{2\,!}+\frac{1}{4\,!}+\frac{1}{6\,!}+.....\infty }{1+\frac{1}{3\,!}+\frac{1}{5\,!}+\frac{1}{7\,!}+.....\infty }=\]
  A)
  \[\frac{e+1}{e-1}\] done clear
  B)
  \[\frac{e-1}{e+1}\] done clear
  C)
  \[\frac{{{e}^{2}}+1}{{{e}^{2}}-1}\] done clear
  D)
  \[\frac{{{e}^{2}}-1}{{{e}^{2}}+1}\] done clear
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• question_answer2)
  \[\frac{1}{1\,!}+\frac{4}{2\,!}+\frac{7}{3\,!}+\frac{10}{4\,!}+.....\infty =\]
  A)
  \[e+4\] done clear
  B)
  \[2+e\] done clear
  C)
  \[3+e\] done clear
  D)
  \[e\] done clear
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• question_answer3)
  In  the expansion of  \[\frac{1+x}{1\,!}+\frac{{{(1+x)}^{2}}}{2\,!}+\frac{{{(1+x)}^{3}}}{3\,!}+.....,\] the coefficient of \[{{x}^{n}}\] will be
  A)
  \[\frac{1}{n\,!}\] done clear
  B)
  \[\frac{1}{n\,!}+\frac{1}{(n+1)\,!}\] done clear
  C)
  \[\frac{e}{n\,!}\] done clear
  D)
  \[e\,\left[ \frac{1}{n\,!}+\frac{1}{(n+1)\,!} \right]\] done clear
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• question_answer4)
  If  \[n\] is even, then in the expansion of  \[{{\left( 1+\frac{{{x}^{2}}}{2\,!}+\frac{{{x}^{4}}}{4\,!}+...... \right)}^{2}}\], the coefficient of \[{{x}^{n}}\] is
  A)
  \[\frac{{{2}^{n}}}{n\,!}\] done clear
  B)
  \[\frac{{{2}^{n}}-2}{n\,\,!}\] done clear
  C)
  \[\frac{{{2}^{n-1}}-1}{n\,!}\] done clear
  D)
  \[\frac{{{2}^{n-1}}}{n\,!}\] done clear
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• question_answer5)
  \[1+\frac{1+2}{1\,!}+\frac{1+2+3}{2\,!}+\frac{1+2+3+4}{3\,!}+....\infty =\]
  A)
  0 done clear
  B)
  1 done clear
  C)
  \[\frac{7e}{2}\] done clear
  D)
  \[2\,e\] done clear
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• question_answer6)
  \[1.5+\frac{2.6}{1\,!}+\frac{3.7}{2\,!}+\frac{4.8}{3\,!}+.....\] is equal to
  A)
  \[13\,e\] done clear
  B)
  \[15\,e\] done clear
  C)
  \[9\,e+1\] done clear
  D)
  \[5\,e\] done clear
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• question_answer7)
  If  \[S=\sum\limits_{n=0}^{\infty }{\frac{{{(\log x)}^{2n}}}{(2n)\,!},}\] then \[S\] =
  A)
  \[x+{{x}^{-1}}\] done clear
  B)
  \[x-{{x}^{-1}}\] done clear
  C)
  \[\frac{1}{2}(x+{{x}^{-1}})\] done clear
  D)
  None of these done clear
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• question_answer8)
  The sum of the series\[\frac{4}{1\,!}+\frac{11}{2\,!}+\frac{22}{3\,!}+\frac{37}{4\,!}+\frac{56}{5\,!}+...\]is  [Kurukshetra CEE 2002]
  A)
  6 e done clear
  B)
  6 e ? 1 done clear
  C)
  5 e done clear
  D)
  5 e + 1 done clear
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• question_answer9)
  In  the expansion of  \[\frac{a+bx+c{{x}^{2}}}{{{e}^{x}}}\], the coefficient of \[{{x}^{n}}\] will be
  A)
  \[\frac{a\,{{(-1)}^{n}}}{n\,!}+\frac{b{{(-1)}^{n-1}}}{(n-1)\,!}+\frac{c{{(-1)}^{n-2}}}{(n-2)\,!}\] done clear
  B)
  \[\frac{a}{n\,!}+\frac{b}{(n-1)\,!}+\frac{c}{(n-2)\,!}\] done clear
  C)
  \[\frac{\,{{(-1)}^{n}}}{n\,!}+\frac{{{(-1)}^{n-1}}}{(n-1)\,!}+\frac{{{(-1)}^{n-2}}}{(n-2)\,!}\] done clear
  D)
  None of these done clear
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• question_answer10)
  The sum of the series \[1+\frac{1}{4.2\,!}+\frac{1}{16.4\,!}+\frac{1}{64.6\,!}+.....\] and inf. is [AIEEE 2005]
  A)
  \[\frac{e-1}{2\sqrt{e}}\] done clear
  B)
  \[\frac{e+1}{2\sqrt{e}}\] done clear
  C)
  \[\frac{e-1}{\sqrt{e}}\] done clear
  D)
  \[\frac{e+1}{\sqrt{e}}\] done clear
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• question_answer11)
  If \[m,\,n\] are the roots of the equation \[{{x}^{2}}-x-1=0\],  then the value of    \[\frac{\left( 1+m{{\log }_{e}}3+\frac{{{(m{{\log }_{e}}3)}^{2}}}{2\,!\,}+...\infty  \right)\,\,\left( 1+n{{\log }_{e}}3+\frac{{{(n{{\log }_{e}}3)}^{2}}}{2\,!\,}+..\infty  \right)\,}{\left( 1+mn{{\log }_{e}}3+\frac{{{(mn{{\log }_{e}}3)}^{2}}}{2\,!}+.....\infty  \right)}\]
  A)
  9 done clear
  B)
  3 done clear
  C)
  0 done clear
  D)
  1 done clear
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• question_answer12)
  \[\frac{1}{3}+\frac{1}{2\,.\,{{3}^{2}}}+\frac{1}{3\,.\,{{3}^{3}}}+\frac{1}{4\,.\,{{3}^{4}}}+.....\infty =\]  [MNR 1975]
  A)
  \[{{\log }_{e}}2-{{\log }_{e}}3\] done clear
  B)
  \[{{\log }_{e}}3-{{\log }_{e}}2\] done clear
  C)
  \[{{\log }_{e}}6\] done clear
  D)
  None of these done clear
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• question_answer13)
  If  \[|x|<1\], then the coefficient of \[{{x}^{5}}\] in the expansion of \[(1-x){{\log }_{e}}(1-x)\] is
  A)
  1/2 done clear
  B)
  1/4 done clear
  C)
  1/20 done clear
  D)
  1/10 done clear
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• question_answer14)
  In  the expansion of \[{{\log }_{e}}\frac{1}{1-x-{{x}^{2}}+{{x}^{3}}}\],  the coefficient of \[x\] is
  A)
  0 done clear
  B)
  1 done clear
  C)
  ? 1 done clear
  D)
  1/2 done clear
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• question_answer15)
  The sum of the series \[1+\frac{3}{2\,!}+\frac{7}{3\,!}+\frac{15}{4\,!}+.....\text{to}\,\infty \] is  [Kerala (Engg.) 2005]
  A)
  \[e(e+1)\] done clear
  B)
  \[e\,(1-e)\] done clear
  C)
  \[3e-1\] done clear
  D)
  \[3e\] done clear
  E)
  (e) \[e\,(e-1)\] done clear
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• question_answer16)
  \[1+\frac{2}{3}-\frac{2}{4}+\frac{2}{5}-......\infty =\]
  A)
  \[{{\log }_{e}}3\] done clear
  B)
  \[{{\log }_{e}}4\] done clear
  C)
  \[{{\log }_{e}}\left( \frac{e}{2} \right)\] done clear
  D)
  \[{{\log }_{e}}\left( \frac{2}{3} \right)\] done clear
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• question_answer17)
  The value of  \[{{\log }_{e}}\left( 1+a{{x}^{2}}+{{a}^{2}}+\frac{a}{{{x}^{2}}} \right)\] is
  A)
  \[a\,\left( {{x}^{2}}-\frac{1}{{{x}^{2}}} \right)-\frac{{{a}^{2}}}{2}\,\left( {{x}^{4}}-\frac{1}{{{x}^{4}}} \right)+\frac{{{a}^{3}}}{3}\,\left( {{x}^{6}}-\frac{1}{{{x}^{6}}} \right)-.....\] done clear
  B)
   \[a\,\left( {{x}^{2}}+\frac{1}{{{x}^{2}}} \right)-\frac{{{a}^{2}}}{2}\,\left( {{x}^{4}}+\frac{1}{{{x}^{4}}} \right)+\frac{{{a}^{3}}}{3}\,\left( {{x}^{6}}+\frac{1}{{{x}^{6}}} \right)-.....\] done clear
  C)
  \[a\,\left( {{x}^{2}}+\frac{1}{{{x}^{2}}} \right)+\frac{{{a}^{2}}}{2}\,\left( {{x}^{4}}+\frac{1}{{{x}^{4}}} \right)+\frac{{{a}^{3}}}{3}\,\left( {{x}^{6}}+\frac{1}{{{x}^{6}}} \right)+.....\] done clear
  D)
   \[a\,\left( {{x}^{2}}-\frac{1}{{{x}^{2}}} \right)+\frac{{{a}^{2}}}{2}\,\left( {{x}^{4}}-\frac{1}{{{x}^{4}}} \right)+\frac{{{a}^{3}}}{3}\,\left( {{x}^{6}}-\frac{1}{{{x}^{6}}} \right)+.....\] done clear
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• question_answer18)
  \[{{\log }_{e}}(1+x)=\sum\limits_{i=1}^{\infty }{\left[ \frac{{{(-1)}^{i+1}}{{x}^{i}}}{i} \right]}\] is defined for  [Roorkee 1990]
  A)
  \[x\in (-1,\,1)\] done clear
  B)
  Any positive (+) real x done clear
  C)
  \[x\in (-1,\,1]\] done clear
  D)
  Any positive (+) real \[x(x\ne 1)\] done clear
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• question_answer19)
  If  \[y=2{{x}^{2}}-1\], then  \[\left[ \frac{1}{y}+\frac{1}{3{{y}^{3}}}+\frac{1}{5{{y}^{5}}}+.... \right]\] is equal to
  A)
  \[\frac{1}{2}\left[ \frac{1}{{{x}^{2}}}-\frac{1}{2{{x}^{4}}}+\frac{1}{3{{x}^{6}}}-..... \right]\] done clear
  B)
  \[\frac{1}{2}\left[ \frac{1}{{{x}^{2}}}+\frac{1}{2{{x}^{4}}}+\frac{1}{3{{x}^{6}}}+..... \right]\] done clear
  C)
  \[\frac{1}{2}\left[ \frac{1}{{{x}^{2}}}+\frac{1}{3{{x}^{6}}}+\frac{1}{5{{x}^{10}}}+..... \right]\] done clear
  D)
  \[\frac{1}{2}\left[ \frac{1}{{{x}^{2}}}-\frac{1}{3{{x}^{6}}}+\frac{1}{5{{x}^{10}}}-..... \right]\] done clear
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• question_answer20)
  If  \[x,\,y,z\] are three consecutive positive integers, then \[\frac{1}{2}{{\log }_{e}}x+\frac{1}{2}{{\log }_{e}}z+\frac{1}{2xz+1}+\frac{1}{3}{{\left( \frac{1}{2xz+1} \right)}^{3}}+....=\]
  A)
  \[{{\log }_{e}}x\] done clear
  B)
  \[{{\log }_{e}}y\] done clear
  C)
  \[{{\log }_{e}}z\] done clear
  D)
  None of these done clear
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