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find the sum of n terms of the series 3+8+22+72+266+1036...............

find the sum of n terms of the series 3+8+22+72+266+1036...............

Grade:12

1 Answers

Pratik Tibrewal
askIITians Faculty 37 Points
8 years ago
S = 3 + 8 + 22 + 72 + 266 + 1036 + ... ------1
S = 3 + 8 + 22 + 72 + 266 + ... ------2
Eqn 1 - Eqn 2, we get
0 = 3 + 5 + 14 + 50 + 194 + 770 + ..... - Tn ---3
Tn = 3 + 5 + 14 + 50 + 194 + 770 + .... -------4
Tn = 3 + 5 + 14 + 50 + 194 + ....... ----------5
0 = 3 + 2 + 9 + 36 + 144 + 576 + ..... -tn -------6
tn = 3 + 2 + G.P. with first term 9, common ratio 4, and number of terms are n -2
tn = 5 + 9((4)^(n-2) - 1)/(4-1))
tn = 2 + (3/16) ((4)^(n))
Tn = 3 + SUM(tn) where number of terms are (n-1)
Tn = 3 + 2(n-1) + (3/16) ((4)(4^(n-1) - 1)/(4-1)) -------7
Tn = 3/4 + 2n + (4^(n-2))
Sn = SUM(Tn) where number of terms are 'n'
Sn = 3n/4 + (n)(n+1) + (1/12)((4^n - 1)


Thanks and Regards,
askiitians faculty,
BTech IITG

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