13. al = a2 = 2, an = an-1-1, n > 2

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Solution

The given terms are a 1 = a 2 =2. Also, a n = a n−1 −1. The given condition is n>2.

Substitute the value of n=3 in a n = a n−1 −1.

a 3 = a 2 −1 =2−1 =1

Similarly, substitute the value of n=4 in a n = a n−1 −1.

a 4 = a 3 −1 =1−1 =0

Similarly, substitute the value of n=5 in a n = a n−1 −1.

a 5 = a 4 −1 =0−1 =−1

Thus, the first five terms of the sequence are 2,2,1,0 and −1. The corresponding series is 2+2+1+0+( −1 )+….

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Similar questions

a_{1} = a_{2} = 2, a_{n} = a_{n - 1} - 1, n > 2

Write the first five terms in each of the following sequences :

(i) $a_{1} = 1, a_{n} = a_{n - 1} + 2, n > 1$

(ii) $a_{1} = 1 = a_{2}, a_{n} = a_{n - 1} + a_{n} - 2, n > 2$

(iii) $a_{1} = a_{2} = 2, a_{n} = a_{n - 1} - 1, n > 2$

a_{1}, a_{2}, . . . . a_{n - 1}

n

(1 - a_{1}) (1 - a_{2}) (1 - a_{n - 1}) . . . = n

1 + a_{1} + a_{2} + . . . . + a_{n - 1} = 0

\frac{1}{2 - a_{1}} + \frac{1}{2 - a_{2}} + . . . . \frac{1}{2 - a_{n - 1}} = \frac{(n - 2) 2^{n - 1} + 1}{2^{n} - 1}

1 = a_{1} = a_{2} = 2, a_{n} = a_{n - 1} + a_{n - 2}, n > 2.

\frac{a_{n + 1}}{a_{n}},

n = 1, 2, 3, 4, 5

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