Permutations and Combinations

  • Question 1
    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

    Question 2
    CBSEENMA11012953

    Write the first five terms of the sequence whose nth terms are:

    straight a subscript straight n space equals space left parenthesis negative 1 right parenthesis to the power of straight n minus 1 end exponent space 5 to the power of straight n plus 1 end exponent







    Solution

    Here, straight a subscript straight n space equals space left parenthesis negative 1 right parenthesis to the power of straight n minus 1 end exponent 5 to the power of straight n plus 1 end exponent
    Putting n = 1, 2, 3, 4, 5, we get
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                           straight a subscript 2 space equals space left parenthesis negative 1 right parenthesis to the power of 2 minus 1 end exponent 5 to the power of 2 plus 1 end exponent equals negative 5 cubed equals negative 125
                           straight a subscript 3 space equals space left parenthesis negative 1 right parenthesis to the power of 3 minus 1 end exponent 5 to the power of 3 plus 1 end exponent space equals space 5 to the power of 4 space equals space 625
straight a subscript 4 space equals space left parenthesis negative 1 right parenthesis to the power of 4 minus 1 end exponent 5 to the power of 4 plus 1 end exponent equals space minus 5 to the power of 5 space equals space minus 3125
straight a subscript 5 space equals space left parenthesis negative 1 right parenthesis to the power of 5 minus 1 end exponent 5 to the power of 5 plus 1 end exponent equals space 5 to the power of 6 space equals space 15625

    ∴   First five terms are: 25, -125, 625, -3125, 15625


                           
                          
           
    Question 3
    CBSEENMA11012954

    Find the indicated terms in the following sequence whose nth terms are :

    space space straight a subscript straight n space equals space fraction numerator straight n left parenthesis straight n minus 2 right parenthesis over denominator straight n plus 3 end fraction

    Solution

    Here, straight a subscript straight n space equals space fraction numerator straight n left parenthesis straight n minus 2 right parenthesis over denominator straight n plus 3 end fraction semicolon space space straight a subscript 19
    Putting n = 19, we get straight a subscript 19 space equals space fraction numerator 19 left parenthesis 9 minus 2 right parenthesis over denominator 19 plus 3 end fraction space equals space fraction numerator 19 cross times 17 over denominator 22 end fraction equals 323 over 22

    Question 4
    CBSEENMA11012955

    Find the indicated terms in the following sequence whose nth terms are:

    straight h left parenthesis straight n right parenthesis space equals space fraction numerator straight n squared plus 2 straight n over denominator 2 straight n end fraction semicolon space straight h left parenthesis straight n minus 1 right parenthesis comma space straight h left parenthesis 16 right parenthesis


    Solution

    Here,   straight h left parenthesis straight n right parenthesis space equals space fraction numerator straight n squared plus 2 straight n over denominator 2 straight n end fraction                                      ...(i)
    Replacing n by n-1 in (1), we get
                          straight h left parenthesis straight n minus 1 right parenthesis space equals space fraction numerator left parenthesis straight n minus 1 right parenthesis squared plus 2 left parenthesis straight n minus 1 right parenthesis over denominator 2 left parenthesis straight n minus 1 right parenthesis end fraction space equals space fraction numerator straight n squared minus 2 straight n plus 1 plus 2 straight n minus 2 over denominator 2 left parenthesis straight n minus 1 right parenthesis end fraction space equals fraction numerator straight n squared minus 1 over denominator 2 left parenthesis straight n minus 1 right parenthesis end fraction space equals space fraction numerator straight n plus 1 over denominator 2 end fraction
    Putting n = 16 in (I), we get straight h left parenthesis 16 right parenthesis space equals space fraction numerator 16 squared plus 2 left parenthesis 16 right parenthesis over denominator 2 left parenthesis 16 right parenthesis end fraction space equals space fraction numerator 256 plus 32 over denominator 32 end fraction space equals space 288 over 32 space equals space 9
                         

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