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question_answer1)
The sum of \[(n+1)\] terms of \[\frac{1}{1}+\frac{1}{1+2}+\frac{1}{1+2+3}+......\,\,\text{is }\] [RPET 1999]
A)
\[\frac{n}{n+1}\] done
clear
B)
\[\frac{2n}{n+1}\] done
clear
C)
\[\frac{2}{n\,(n+1)}\] done
clear
D)
\[\frac{2\,(n+1)}{n+2}\] done
clear
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question_answer2)
The sum of \[(n-1)\] terms of \[1+(1+3)+\] \[(1+3+5)+.......\] is [RPET 1999]
A)
\[\frac{n\,(n+1)\,(2n+1)}{6}\] done
clear
B)
\[\frac{{{n}^{2}}(n+1)}{4}\] done
clear
C)
\[\frac{n\,(n-1)\,(2n-1)}{6}\] done
clear
D)
\[{{n}^{2}}\] done
clear
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question_answer3)
The sum of first \[n\] terms of the given series \[{{1}^{2}}+{{2.2}^{2}}+{{3}^{2}}+{{2.4}^{2}}+{{5}^{2}}+{{2.6}^{2}}+............\]is\[\frac{n{{(n+1)}^{2}}}{2}\], when \[n\] is even. When \[n\] is odd, the sum will be [IIT 1988]
A)
\[\frac{n{{(n+1)}^{2}}}{2}\] done
clear
B)
\[\frac{1}{2}{{n}^{2}}(n+1)\] done
clear
C)
\[n{{(n+1)}^{2}}\] done
clear
D)
None of these done
clear
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question_answer4)
The sum to \[n\] terms of the series \[{{2}^{2}}+{{4}^{2}}+{{6}^{2}}+...........\] is [MP PET 1994]
A)
\[\frac{n(n+1)(2n+1)}{3}\] done
clear
B)
\[\frac{2n(n+1)(2n+1)}{3}\] done
clear
C)
\[\frac{n(n+1)(2n+1)}{6}\] done
clear
D)
\[\frac{n(n+1)(2n+1)}{9}\] done
clear
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question_answer5)
The sum of the series \[1+(1+2)+(1+2+3)+............\]upto \[n\] terms, will be [MP PET 1986]
A)
\[{{n}^{2}}-2n+6\] done
clear
B)
\[\frac{n(n+1)(2n-1)}{6}\] done
clear
C)
\[{{n}^{2}}+2n+6\] done
clear
D)
\[\frac{n(n+1)(n+2)}{6}\] done
clear
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question_answer6)
The sum of \[n\] terms of the series whose \[{{n}^{hth}}\] term is \[n(n+1)\] is equal to
A)
\[\frac{n(n+1)(n+2)}{3}\] done
clear
B)
\[\frac{(n+1)(n+2)(n+3)}{12}\] done
clear
C)
\[{{n}^{2}}(n+2)\] done
clear
D)
\[n(n+1)(n+2)\] done
clear
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question_answer7)
The sum 1(1!) + 2(2!) + 3(3!) + ....+n (n!) equals [AMU 1999; DCE 2005]
A)
\[3\,(n\,!)\,+\,n-3\] done
clear
B)
\[(n+1)!\,-\,(n-1)!\] done
clear
C)
\[(n+1)\,!\,-1\] done
clear
D)
\[2\,(n\,!)-2n-1\] done
clear
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question_answer8)
\[\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+........+.......\frac{1}{n.(n+1)}\]equals [AMU 1995; RPET 1996; UPSEAT 1999, 2001]
A)
\[\frac{1}{n(n+1)}\] done
clear
B)
\[\frac{n}{n+1}\] done
clear
C)
\[\frac{2n}{n+1}\] done
clear
D)
\[\frac{2}{n(n+1)}\] done
clear
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question_answer9)
The sum of the series \[3.6+4.7+5.8+........\]upto \[(n-2)\] terms [EAMCET 1980]
A)
\[{{n}^{3}}+{{n}^{2}}+n+2\] done
clear
B)
\[\frac{1}{6}(2{{n}^{3}}+12{{n}^{2}}+10n-84)\] done
clear
C)
\[{{n}^{3}}+{{n}^{2}}+n\] done
clear
D)
None of these done
clear
View Solution play_arrow
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question_answer10)
If \[\sum\limits_{i=1}^{n}{i=\frac{n(n+1)}{2}}\], then \[\sum\limits_{i=1}^{n}{(3i-2)=}\]
A)
\[\frac{n(3n-1)}{2}\] done
clear
B)
\[\frac{n(3n+1)}{2}\] done
clear
C)
\[n(3n+2)\] done
clear
D)
\[\frac{n(3n+1)}{4}\] done
clear
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question_answer11)
The sum of the series \[{{1}^{2}}.2+{{2}^{2}}.3+{{3}^{2}}.4+........\] to n terms is
A)
\[\frac{{{n}^{3}}{{(n+1)}^{3}}(2n+1)}{24}\] done
clear
B)
\[\frac{n(n+1)(3{{n}^{2}}+7n+2)}{12}\] done
clear
C)
\[\frac{n(n+1)}{6}[n(n+1)+(2n+1)]\] done
clear
D)
\[\frac{n(n+1)}{12}[6n(n+1)+2(2n+1)]\] done
clear
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question_answer12)
The sum of the series \[1.2.3+2.3.4+3.4.5+.......\] to n terms is [Kurukshetra CEE 1998]
A)
\[n(n+1)(n+2)\] done
clear
B)
\[(n+1)(n+2)(n+3)\] done
clear
C)
\[\frac{1}{4}n(n+1)(n+2)(n+3)\] done
clear
D)
\[\frac{1}{4}(n+1)(n+2)(n+3)\] done
clear
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question_answer13)
The sum of \[{{1}^{3}}+{{2}^{3}}+{{3}^{3}}+{{4}^{3}}+.....+{{15}^{3}}\],is [MP PET 2003]
A)
22000 done
clear
B)
10,000 done
clear
C)
14,400 done
clear
D)
15,000 done
clear
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question_answer14)
Sum of the squares of first \[n\] natural numbers exceeds their sum by 330, then \[n=\] [Karnataka CET 1998]
A)
8 done
clear
B)
10 done
clear
C)
15 done
clear
D)
20 done
clear
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question_answer15)
Sum of first \[n\] terms in the following series \[{{\cot }^{-1}}3+{{\cot }^{-1}}7+{{\cot }^{-1}}13+{{\cot }^{-1}}21+.............\] is given by
A)
\[{{\tan }^{-1}}\left( \frac{n}{n+2} \right)\] done
clear
B)
\[{{\cot }^{-1}}\left( \frac{n+2}{n} \right)\] done
clear
C)
\[{{\tan }^{-1}}(n+1)-{{\tan }^{-1}}1\] done
clear
D)
All of these done
clear
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question_answer16)
The natural numbers are written as follows \[1\] x\[\begin{matrix} 2 & 3 \\ \end{matrix}\]\[\begin{matrix} 4 & 5 & 6 \\ \end{matrix}\]\[\begin{matrix} 7 & 8 & 9 & 10 \\ . & . & . & . \\ . & . & . & . \\ . & . & . & . \\ \end{matrix}\] The sum of the numbers in the \[{{n}^{th}}\] row is
A)
\[\frac{n}{2}({{n}^{2}}-1)\] done
clear
B)
\[\frac{n}{2}({{n}^{2}}+1)\] done
clear
C)
\[\frac{2}{n}({{n}^{2}}+1)\] done
clear
D)
\[\frac{2}{n}({{n}^{2}}-1)\] done
clear
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question_answer17)
The \[{{n}^{th}}\] term of series \[\frac{1}{1}+\frac{1+2}{2}+\frac{1+2+3}{3}+.......\] will be [AMU 1982]
A)
\[\frac{n+1}{2}\] done
clear
B)
\[\frac{n-1}{2}\] done
clear
C)
\[\frac{{{n}^{2}}+1}{2}\] done
clear
D)
\[\frac{{{n}^{2}}-1}{2}\] done
clear
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question_answer18)
The sum of all the products of the first \[n\] natural numbers taken two at a time is
A)
\[\frac{1}{24}n(n-1)(n+1)(3n+2)\] done
clear
B)
\[\frac{{{n}^{2}}}{48}(n-1)(n-2)\] done
clear
C)
\[\frac{1}{6}n(n+1)(n+2)(n+5)\] done
clear
D)
None of these done
clear
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question_answer19)
The sum of the series \[{{1.3}^{2}}+{{2.5}^{2}}+{{3.7}^{2}}+..........\]upto \[20\] terms is [IIT 1973]
A)
188090 done
clear
B)
189080 done
clear
C)
199080 done
clear
D)
None of these done
clear
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question_answer20)
\[\frac{{{1}^{3}}+{{2}^{3}}+{{3}^{3}}+{{4}^{3}}+{{........12}^{3}}}{{{1}^{2}}+{{2}^{2}}+{{3}^{2}}+{{4}^{2}}+{{.........12}^{2}}}=\] [MP PET 1998]
A)
\[\frac{234}{25}\] done
clear
B)
\[\frac{243}{35}\] done
clear
C)
\[\frac{263}{27}\] done
clear
D)
None of these done
clear
View Solution play_arrow
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question_answer21)
Sum of the \[n\] terms of the series \[\frac{3}{{{1}^{2}}}+\frac{5}{{{1}^{2}}+{{2}^{2}}}+\frac{7}{{{1}^{2}}+{{2}^{2}}+{{3}^{2}}}\,+...\,\,\text{is}\] [Pb. CET 1999; RPET 2001]
A)
\[\frac{2n}{n+1}\] done
clear
B)
\[\frac{4n}{n+1}\] done
clear
C)
\[\frac{6n}{n+1}\] done
clear
D)
\[\frac{9n}{n+1}\] done
clear
View Solution play_arrow
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question_answer22)
The sum of the series \[1.3.5+.2.5.8+3.7.11+.........\]upto \['n'\] terms is [Dhanbad Engg. 1972]
A)
\[\frac{n\,(n+1)(9{{n}^{2}}+23n+13)}{6}\] done
clear
B)
\[\frac{n\,(n-1)(9{{n}^{2}}+23n+12)}{6}\] done
clear
C)
\[\frac{(n+1)(9{{n}^{2}}+23n+13)}{6}\] done
clear
D)
\[\frac{n\,(9{{n}^{2}}+23n+13)}{6}\] done
clear
View Solution play_arrow
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question_answer23)
Sum of the series \[\frac{2}{3}+\frac{8}{9}+\frac{26}{27}+\frac{80}{81}+.....\] to n terms is [Karnataka CET 2001]
A)
\[n-\frac{1}{2}({{3}^{n}}-1)\] done
clear
B)
\[n+\frac{1}{2}({{3}^{n}}-1)\] done
clear
C)
\[n+\frac{1}{2}(1-{{3}^{-n}})\] done
clear
D)
\[n+\frac{1}{2}({{3}^{-n}}-1)\] done
clear
View Solution play_arrow
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question_answer24)
\[\sum\limits_{m=1}^{n}{{{m}^{2}}}\] is equal to [RPET 1995]
A)
\[\frac{m(m+1)}{2}\] done
clear
B)
\[\frac{m(m+1)(2m+1)}{6}\] done
clear
C)
\[\frac{n(n+1)(2n+1)}{6}\] done
clear
D)
\[\frac{n(n+1)}{2}\] done
clear
View Solution play_arrow
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question_answer25)
The sum of \[n\] terms of the following series \[1.2+2.3+3.4+4.5+.........\] shall be [MNR 1980]
A)
\[{{n}^{3}}\] done
clear
B)
\[\frac{1}{3}n\,(n+1)(n+2)\] done
clear
C)
\[\frac{1}{6}n\,(n+1)(n+2)\] done
clear
D)
\[\frac{1}{3}n\,(n+1)(2n+1)\] done
clear
View Solution play_arrow
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question_answer26)
\[{{11}^{3}}+{{12}^{3}}+....+{{20}^{3}}\] [Pb. CET 1997; RPET 2002]
A)
Is divisible by 5 done
clear
B)
Is an odd integer divisible by 5 done
clear
C)
Is an even integer which is not divisible by 5 done
clear
D)
Is an odd integer which is not divisible by 5 done
clear
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question_answer27)
The sum to \[n\] terms of the infinite series \[{{1.3}^{2}}+{{2.5}^{2}}+{{3.7}^{2}}+..........\infty \] is [AMU 1982]
A)
\[\frac{n}{6}(n+1)(6{{n}^{2}}+14n+7)\] done
clear
B)
\[\frac{n}{6}(n+1)(2n+1)(3n+1)\] done
clear
C)
\[4{{n}^{3}}+4{{n}^{2}}+n\] done
clear
D)
None of these done
clear
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question_answer28)
If the \[{{n}^{th}}\] term of a series be \[3+n\,(n-1)\], then the sum of \[n\] terms of the series is
A)
\[\frac{{{n}^{2}}+n}{3}\] done
clear
B)
\[\frac{{{n}^{3}}+8n}{3}\] done
clear
C)
\[\frac{{{n}^{2}}+8n}{5}\] done
clear
D)
\[\frac{{{n}^{2}}-8n}{3}\] done
clear
View Solution play_arrow
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question_answer29)
The sum to \[n\] terms of\[(2n-1)+2\,(2n-3)\] \[+3\,(2n-5)+.....\] is [AMU 2001]
A)
\[(n+1)\,(n+2)\,(n+3)/6\] done
clear
B)
\[n\,(n+1)\,(n+2)/6\] done
clear
C)
\[n\,(n+1)\,(2n+3)\,\] done
clear
D)
\[n\,(n+1)\,(2n+1)/6\] done
clear
View Solution play_arrow
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question_answer30)
First term of the \[{{11}^{th}}\]group in the following groups (1), (2, 3, 4), (5, 6, 7, 8, 9),???.is
A)
89 done
clear
B)
97 done
clear
C)
101 done
clear
D)
123 done
clear
View Solution play_arrow
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question_answer31)
\[{{11}^{2}}+{{12}^{2}}+{{13}^{2}}+{{.......20}^{2}}=\] [MP PET 1995]
A)
2481 done
clear
B)
2483 done
clear
C)
2485 done
clear
D)
2487 done
clear
View Solution play_arrow
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question_answer32)
The sum to infinity of the following series \[\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...........\] shall be [MNR 1982]
A)
\[\infty \] done
clear
B)
1 done
clear
C)
0 done
clear
D)
None of these done
clear
View Solution play_arrow
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question_answer33)
\[\frac{\frac{1}{2}.\frac{2}{2}}{{{1}^{3}}}+\frac{\frac{2}{2}.\frac{3}{2}}{{{1}^{3}}+{{2}^{3}}}+\frac{\frac{3}{2}.\frac{4}{2}}{{{1}^{3}}+{{2}^{3}}+{{3}^{3}}}+.....n\] terms = [EAMCET 2000]
A)
\[{{\left( \frac{n}{n+1} \right)}^{2}}\] done
clear
B)
\[{{\left( \frac{n}{n+1} \right)}^{3}}\] done
clear
C)
\[\left( \frac{n}{n+1} \right)\] done
clear
D)
\[\left( \frac{1}{n+1} \right)\] done
clear
View Solution play_arrow
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question_answer34)
If the sum of \[1+\frac{1+2}{2}+\frac{1+2+3}{3}+.....\] to n terms is S, then S is equal to [Kerala (Engg.) 2002]
A)
\[\frac{n(n+3)}{4}\] done
clear
B)
\[\frac{n(n+2)}{4}\] done
clear
C)
\[\frac{n(n+1)\,(n+2)}{6}\] done
clear
D)
\[{{n}^{2}}\] done
clear
View Solution play_arrow
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question_answer35)
The nth term of the series \[\frac{2}{1!}+\frac{7}{2\,!}+\frac{15}{3\,!}+\frac{26}{4\,!}+.....\] is
A)
\[\frac{n\,(3n-1)}{2(n)\,!}\] done
clear
B)
\[\frac{n\,(3n+1)}{2\,(n)\,!}\] done
clear
C)
\[\frac{n}{2}\frac{3n}{(n)\,!}\] done
clear
D)
None of these done
clear
View Solution play_arrow
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question_answer36)
If \[{{t}_{n}}=\frac{1}{4}(n+2)\,(n+3)\] for n = 1, 2, 3,?? then \[\frac{1}{{{t}_{1}}}+\frac{1}{{{t}_{2}}}+\frac{1}{{{t}_{3}}}+....+\frac{1}{{{t}_{2003}}}=\] [EAMCET 2003]
A)
\[\frac{4006}{3006}\] done
clear
B)
\[\frac{4003}{3007}\] done
clear
C)
\[\frac{4006}{3008}\] done
clear
D)
\[\frac{4006}{3009}\] done
clear
View Solution play_arrow
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question_answer37)
If \[\frac{1}{{{1}^{4}}}+\frac{1}{{{2}^{4}}}+\frac{1}{{{3}^{4}}}+.....+\infty =\frac{{{\pi }^{4}}}{90}\], then the value of \[\frac{1}{{{1}^{4}}}+\frac{1}{{{3}^{4}}}+\frac{1}{{{5}^{4}}}+.....\infty \]is [AMU 2005]
A)
\[\frac{{{\pi }^{4}}}{96}\] done
clear
B)
\[\frac{{{\pi }^{4}}}{45}\] done
clear
C)
\[\frac{89}{90}{{\pi }^{4}}\] done
clear
D)
None of these done
clear
View Solution play_arrow
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question_answer38)
The value of \[\frac{1}{(1+a)(2+a)}+\frac{1}{(2+a)(3+a)}\] \[\frac{1}{(3+a)(4+a)}\]+ ..... +\[\infty \] is, (where a is a constant) [AMU 2005]
A)
\[\frac{1}{1+a}\] done
clear
B)
\[\frac{2}{1+a}\] done
clear
C)
\[\infty \] done
clear
D)
None of these done
clear
View Solution play_arrow