The Fibonacci sequence is defined by              

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 Multiple Choice QuestionsShort Answer Type

1.

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

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2.

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







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3.

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

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4.

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


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5.

The Fibonacci sequence is defined by

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#6 {main}</pre>

Find fraction numerator straight a subscript straight n plus 1 over denominator straight a subscript straight n end fraction, for n = 1, 2, 3, 4, 5.



The Fibonacci sequence is given by

                              straight a subscript 1 space equals space straight a subscript 2 space equals space 1

and                          straight a subscript straight n space equals space straight a subscript straight n minus 1 end subscript plus straight a subscript straight n minus 2 end subscript

Putting n =3 in (i), we get straight a subscript 3 space equals space straight a subscript 3 minus 1 end subscript space plus space straight a subscript 3 minus 2 end subscript space equals space straight a subscript 2 space plus space straight a subscript 1 space equals space 1 space plus space 1 space equals space 2

Putting n = 4 in (i), we get straight a subscript 4 space equals space straight a subscript 4 minus 1 end subscript space plus space straight a subscript 4 minus 2 end subscript space equals space straight a subscript 3 plus straight a subscript 2 space equals space 2 space plus space 1 space equals 3

Putting n = 5 in (i), we get straight a subscript 5 equals straight a subscript 5 minus 1 end subscript plus straight a subscript 5 minus 2 end subscript equals straight a subscript 4 plus straight a subscript 3 equals 3 plus 2 equals 5

Putting n = 6 in (i), we get straight a subscript 6 space equals space straight a subscript 6 minus 1 end subscript space plus space straight a subscript 6 minus 2 end subscript space equals space straight a subscript 5 plus straight a subscript 4 space equals space 5 space plus space 3 space equals 8

Now, let       straight t subscript straight n space equals space straight a subscript straight n plus 1 end subscript over straight a subscript straight n

                  straight t subscript 1 space equals space fraction numerator straight a subscript 1 plus 1 over denominator straight a subscript 1 end fraction space equals space fraction numerator straight a subscript 2 over denominator straight a subscript 1 space end fraction space equals space 1 over 1 space equals space 1 commastraight t subscript 2 space equals space fraction numerator straight a subscript 2 plus 1 over denominator straight a subscript 2 end fraction space equals space straight a subscript 3 over straight a subscript 2 space equals space 2 over 1 equals 2 comma space space space straight t subscript 3 space equals space fraction numerator straight a subscript 3 plus 1 over denominator straight a subscript 3 end fraction space equals space straight a subscript 4 over straight a subscript 3 equals space 3 over 2

                 straight t subscript 4 space equals space fraction numerator straight a subscript 4 plus 1 over denominator straight a subscript 4 end fraction space equals space straight a subscript 5 over straight a subscript 4 space equals space 5 over 3 comma space space straight t subscript 5 space equals space fraction numerator straight a subscript 5 plus 1 over denominator straight a subscript 5 end fraction space equals space straight a subscript 6 over straight a subscript 5 space equals space 8 over 5

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6.

Find the first six terms of the sequence whose first term is 1 and whose (n+l)th term is obtained by adding n to the nth term.

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 Multiple Choice QuestionsLong Answer Type

7.

Consider the sequence defined by tn = an2 + bn + c. If t2 = 3, t4 = 13 and t7 = 113, show that 3tn = 17n2 – 87n + 115

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 Multiple Choice QuestionsShort Answer Type

8.

Write first five terms of the following sequence and obtain the corresponding series.

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#6 {main}</pre> for all n>1

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9.

Write first five terms of the following sequence and obtain the corresponding series.

straight a subscript 1 space equals space straight a subscript 2 space equals space 2 comma space space space straight a subscript straight n space equals space straight a subscript straight n minus 1 end subscript space minus space 1, n>2

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10.

Show that the sequence space space open curly brackets straight t subscript straight n close curly brackets defined by straight t subscript straight n space equals space nA space plus space straight B (where A and B are constant) is an A.P. with common difference A.

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