Question

For the arithmetico-geometric sequence 1,2*3,3*9,4*27,5*81,...tn=?

Answer 1

Answer:

tn = n* (3)^(n-1)

Step-by-step explanation:

tn = n* (3)^(n-1)

Answer 2

The numbers 1, 2, 3, 4, ... form an A.P. whose first term is a 1 and the common difference is d = 1.

Hence, the nth term of this A.P. is a + (n − 1)d = 1 + (n − 1)1 = n - -

The numbers 1, 3, 9, 27, 81, ... form the G.P. whose first term is A = 1 and common ratio is r = 3.

Hence, the nth term of this G.P. is rn-1 = 3n-1

The terms of the given sequence are obtained by multiplying the corresponding terms of the above A.P. and G.P. Hence, the given series is an arithmetico-geometric series whose nth term is

[a + (n - 1) d] rn − 1 = n.3n-1

The sum of first n terms of the series is

Sn = 1+ 2 × 3 + 3 x 9 + 4 × 27+ 5 × 81 + + (n - 1) -3n-2 + n.3n-1 ... (1) -

3.Sn = 1 x 3 + 2x 9 + 3 x 27 + 4 × 81 + + (n - 1) .3n − 1 + n.3n ... (2)

Subtracting (2) from (1), we get,

Sn 3.Sn = 1 + 1 × 3+1 × 9 + 1 × 27 + 1

x 81 + ... + 3n-1 - n.3n

- 2.Sn = 1+ (3 + 32 +33 +34 + ... + 3n

1) - n.3n

= 1 + 3 (1 + 3 + 32 + 33 + ... + 3n-2) - n.3n

1-3⁰-1 =1+3 n.3⁰ - 1 3 :)

n-1) - n.3n 3 = 1 1 - n - 2

n.3⁰ Sn 1 3 + 2 (1 1 - 3⁰ 3n-¹) + 2

n.3⁰ - 1 3 - 3n + 4 2