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question_answer1)
If \[|x|\,<1\], then the sum of the series \[1+2x+3{{x}^{2}}+4{{x}^{3}}+...........\infty \] will be
A)
\[\frac{1}{1-x}\] done
clear
B)
\[\frac{1}{1+x}\] done
clear
C)
\[\frac{1}{{{(1+x)}^{2}}}\] done
clear
D)
\[\frac{1}{{{(1-x)}^{2}}}\] done
clear
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question_answer2)
\[1+\frac{3}{2}+\frac{5}{{{2}^{2}}}+\frac{7}{{{2}^{3}}}+......\,\infty \,\]is equal to [DCE 1999]
A)
3 done
clear
B)
6 done
clear
C)
9 done
clear
D)
12 done
clear
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question_answer3)
The sum of infinite terms of the following series \[1+\frac{4}{5}+\frac{7}{{{5}^{2}}}+\frac{10}{{{5}^{3}}}+.........\] will be [MP PET 1981; RPET 1997; Roorkee 1992; DCE 1996, 2000]
A)
\[\frac{3}{16}\] done
clear
B)
\[\frac{35}{8}\] done
clear
C)
\[\frac{35}{4}\] done
clear
D)
\[\frac{35}{16}\] done
clear
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question_answer4)
The sum of the series \[1+3x+6{{x}^{2}}+10{{x}^{3}}+........\infty \] will be
A)
\[\frac{1}{{{(1-x)}^{2}}}\] done
clear
B)
\[\frac{1}{1-x}\] done
clear
C)
\[\frac{1}{{{(1+x)}^{2}}}\] done
clear
D)
\[\frac{1}{{{(1-x)}^{3}}}\] done
clear
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question_answer5)
\[1+3+7+15+31+..........\]to \[n\] terms = [IIT 1963]
A)
\[{{2}^{n+1}}-n\] done
clear
B)
\[{{2}^{n+1}}-n-2\] done
clear
C)
\[{{2}^{n}}-n-2\] done
clear
D)
None of these done
clear
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question_answer6)
\[2+4+7+11+16+......\]to \[n\] terms = [Roorkee 1977]
A)
\[\frac{1}{6}({{n}^{2}}+3n+8)\] done
clear
B)
\[\frac{n}{6}({{n}^{2}}+3n+8)\] done
clear
C)
\[\frac{1}{6}({{n}^{2}}-3n+8)\] done
clear
D)
\[\frac{n}{6}({{n}^{2}}-3n+8)\] done
clear
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question_answer7)
Sum of n terms of series \[12+16+24+40+.....\] will be [UPSEAT 1999]
A)
\[2\,({{2}^{n}}-1)+8n\] done
clear
B)
\[2({{2}^{n}}-1)+6n\] done
clear
C)
\[3({{2}^{n}}-1)+8n\] done
clear
D)
\[4({{2}^{n}}-1)+8n\] done
clear
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question_answer8)
If \[a,\ b,\ c\] are in H.P., then which one of the following is true [MNR 1985]
A)
\[\frac{1}{b-a}+\frac{1}{b-c}=\frac{1}{b}\] done
clear
B)
\[\frac{ac}{a+c}=b\] done
clear
C)
\[\frac{b+a}{b-a}+\frac{b+c}{b-c}=1\] done
clear
D)
None of these done
clear
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question_answer9)
The sum of the series \[1+\frac{1.3}{6}+\frac{1.3.5}{6.8}+....\infty \]is [UPSEAT 2001]
A)
1 done
clear
B)
0 done
clear
C)
\[\infty \] done
clear
D)
4 done
clear
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question_answer10)
\[{{n}^{th}}\] term of the series \[2+4+7+11+.......\]will be [Roorkee 1977]
A)
\[\frac{{{n}^{2}}+n+1}{2}\] done
clear
B)
\[{{n}^{2}}+n+2\] done
clear
C)
\[\frac{{{n}^{2}}+n+2}{2}\] done
clear
D)
\[\frac{{{n}^{2}}+2n+2}{2}\] done
clear
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question_answer11)
The sum of the series \[1+2x+3{{x}^{2}}+4{{x}^{3}}+.........\]upto \[n\] terms is
A)
\[\frac{1-(n+1){{x}^{n}}+n{{x}^{n+1}}}{{{(1-x)}^{2}}}\] done
clear
B)
\[\frac{1-{{x}^{n}}}{1-x}\] done
clear
C)
\[{{x}^{n+1}}\] done
clear
D)
None of these done
clear
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question_answer12)
The sum of the first \[n\] terms of the series \[\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+.........\] is [IIT 1988; MP PET 1996; RPET 1996, 2000; Pb. CET 1994; DCE 1995, 96]
A)
\[{{2}^{n}}-n-1\] done
clear
B)
\[1-{{2}^{-n}}\] done
clear
C)
\[n+{{2}^{-n}}-1\] done
clear
D)
\[{{2}^{n}}-1\] done
clear
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question_answer13)
The sum of \[1+\frac{2}{5}+\frac{3}{{{5}^{2}}}+\frac{4}{{{5}^{3}}}+...........\]upto \[n\] terms is [MP PET 1982]
A)
\[\frac{25}{16}-\frac{4n+5}{16\times {{5}^{n-1}}}\] done
clear
B)
\[\frac{3}{4}-\frac{2n+5}{16\times {{5}^{n+1}}}\] done
clear
C)
\[\frac{3}{7}-\frac{3n+5}{16\times {{5}^{n-1}}}\] done
clear
D)
\[\frac{1}{2}-\frac{5n+1}{3\times {{5}^{n+2}}}\] done
clear
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question_answer14)
\[{{2}^{1/4}}{{.4}^{1/8}}{{.8}^{1/16}}{{.16}^{1/32}}..........\]is equal to [MNR 1984; MP PET 1998; AIEEE 2002]
A)
1 done
clear
B)
2 done
clear
C)
\[\frac{3}{2}\] done
clear
D)
\[\frac{5}{2}\] done
clear
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question_answer15)
The sum of \[i-2-3i+4+.........\]upto 100 terms, where \[i=\sqrt{-1}\] is
A)
\[50(1-i)\] done
clear
B)
\[25i\] done
clear
C)
\[25(1+i)\] done
clear
D)
\[100(1-i)\] done
clear
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question_answer16)
\[{{99}^{th}}\] term of the series \[2+7+14+23+34+.....\] is [Pb. CET 2003]
A)
9998 done
clear
B)
9999 done
clear
C)
10000 done
clear
D)
100000 done
clear
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question_answer17)
If the set of natural numbers is partitioned into subsets \[{{S}_{1}}=\left\{ 1 \right\},\ {{S}_{2}}=\left\{ 2,\ 3 \right\},\ {{S}_{3}}=\left\{ 4,\ 5,\ 6 \right\}\] and so on. Then the sum of the terms in \[{{S}_{50}}\] is
A)
62525 done
clear
B)
25625 done
clear
C)
62500 done
clear
D)
None of these done
clear
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