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Permutations And Combinations

Question
CBSEENMA11012959

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

Solution

Here,                     straight t subscript straight n space equals space an squared plus bn plus straight c                              ...(i)
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#6 {main}</pre>
rightwards double arrow          4a + 2b + c = 3                                               ...(ii)
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#6 {main}</pre>
rightwards double arrow           16a + 4b + c = 13                                          ...(iii)
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#6 {main}</pre>
rightwards double arrow           49a + 7b + c = 113                                         ...(iv)
Subtracting (iii) from (iv), we get   33a + 3b = 100                ...(v)
Subtracting (ii) from (iii), we get    12a + 2b = 10
rightwards double arrow              6a + b = 5                                                  ...(vi)
Multiplying both sides of (vi) by 3, we get
                 18a + 3b = 15                                               ...(vii)
Subtracting (vii) from (v), we get
                 15 straight a space equals space 85 space space space space space space space space space space space space space rightwards double arrow space space space space space space space space space space straight a space equals space 17 over 3
Using this value of a in (vi), we get
           6 open parentheses 17 over 3 close parentheses space plus space straight b space equals space 5        rightwards double arrow space space space 34 space plus space straight b space equals space 5 space space rightwards double arrow space space space straight b space equals space minus 29
Using values of a and b in (ii), we get
4 open parentheses 17 over 3 close parentheses plus 2 left parenthesis negative 29 right parenthesis plus straight c equals 3 space space space space rightwards double arrow space space straight c space equals space 3 space plus space 58 space minus 68 over 3 space equals space 61 minus 68 over 3 equals fraction numerator 183 minus 68 over denominator 3 end fraction space rightwards double arrow straight c equals space 115 over 3
Using values of a, b, c in (i), we get
             straight t subscript straight n space equals space 17 over 3 straight n squared minus 29 straight n space plus space 115 over 3 space rightwards double arrow space space 3 straight t subscript straight n space equals space 17 straight n squared minus 87 straight n plus 115

Some More Questions From Permutations and Combinations Chapter

Determine K, so that K + 2, 4K – 6 and 3K – 2 are three consecutive terms of an A.P.