JEE Main & Advanced Mathematics Sequence & Series Question Bank

done nth term of special series, Sum to n terms and Infinite number of terms Total Questions - 38

• 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
  View Solution play_arrow

• 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
  View Solution play_arrow

• 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
  View Solution play_arrow

• 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
  View Solution play_arrow

• 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
  View Solution play_arrow

• 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
  View Solution play_arrow

• 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
  View Solution play_arrow

• 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
  View Solution play_arrow

• 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

• 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
  View Solution play_arrow

• 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
  View Solution play_arrow

• 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
  View Solution play_arrow

• 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
  View Solution play_arrow

• 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
  View Solution play_arrow

• 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
  View Solution play_arrow

• 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
  View Solution play_arrow

• 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
  View Solution play_arrow

• 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
  View Solution play_arrow

• 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
  View Solution play_arrow

• 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

• 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

• 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

• 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

• 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

• 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

• 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
  View Solution play_arrow

• 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
  View Solution play_arrow

• 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

• 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

• 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

• 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

• 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

• 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

• 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

• 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

• 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

• 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

• 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