Solved papers for JEE Main & Advanced AIEEE Solved Paper-2004

done AIEEE Solved Paper-2004 Total Questions - 2

• question_answer1) The sum of the first\[n\]terms of the 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 is
  A)
  \[\frac{3n(n+1)}{2}\]     
  done clear
  B)
         \[\frac{{{n}^{2}}(n+1)}{2}\]        
  done clear
  C)
         \[\frac{n{{(n+1)}^{2}}}{4}\]        
  done clear
  D)
         \[{{\left[ \frac{n(n+1)}{4} \right]}^{2}}\]
  done clear
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• question_answer2) The sum of series\[\frac{1}{2!}+\frac{1}{4!}+\frac{1}{6!}+.....\]is
  A)
  \[\frac{({{e}^{2}}-1)}{2}\]
  done clear
  B)
                         \[\frac{{{(e-1)}^{2}}}{2e}\]         
  done clear
  C)
         \[\frac{({{e}^{2}}-1)}{2e}\]         
  done clear
  D)
         \[\frac{({{e}^{2}}-2)}{e}\]
  done clear
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