Permutations and Combinations

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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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Question
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
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
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