Permutations And Combinations

Question
CBSEENMA11012952

Write the first five terms of each of the sequences whose nth terms are :

straight a subscript straight n space equals space fraction numerator straight n left parenthesis straight n squared plus 5 right parenthesis over denominator 4 end fraction

Solution

Here straight a subscript straight n space equals space fraction numerator straight n left parenthesis straight n squared plus 5 right parenthesis over denominator 4 end fraction
Putting n = 1, 2, 3, 4, 5, we get
        space space straight a subscript 1 space equals space fraction numerator 1 left parenthesis 1 squared plus 5 right parenthesis over denominator 4 end fraction space equals space 6 over 4 space equals space 3 over 2 comma space space space straight a subscript 2 space equals space fraction numerator 2 left parenthesis 2 squared plus 5 right parenthesis over denominator 4 end fraction space equals space 9 over 2 comma space space straight a subscript 3 space equals space fraction numerator 3 left parenthesis 3 squared plus 5 right parenthesis over denominator 4 end fraction space equals space 21 over 2
          straight a subscript 4 equals space fraction numerator 4 left parenthesis 4 squared plus 5 right parenthesis over denominator 4 end fraction equals 21 comma space space space straight a subscript 5 equals space fraction numerator 5 left parenthesis 5 squared plus 5 right parenthesis over denominator 4 end fraction space equals space 75 over 2

∴ First five terms are 3 over 2 comma space space 9 over 2 comma space 21 over 2 comma space 21 comma space 75 over 2

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