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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