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DSSSB TGT Maths Female Subject Concerned - 22 Sep 2018 Shift 2

Option 3 : \(\dfrac{n(n-1)}{2}\)

__Concept:__

**Recurrence Relation:**

A recurrence relation relates the nth term of a sequence to its predecessors. These relations are related to recursive algorithms.

**Definition:**

A recurrence relation for a sequence a_{0}, a_{1}, a_{2},.... is a formula (equation) that relates each term a_{n} to certain of its predecessors a0, a1, a2,...., a_{n-1}. The initial conditions for such a recurrence relation specify the values of a0, a1, a2,...., an-1. For example, the recursive formula for the sequence 3, 8, 13, 18, 23 is

a_{1} = 3, a_{n} = a_{n-1} + 1, \(2\leq n<\infty\)

__Calculation:__

__Given:__

f(n) = (n - 1) + f(n - 1), n > 72, f(2) = 1

put, n = 3

f(3) = (3 - 1) + f(3 - 1) = 2 + f(2) = 2 + 1 = 3

put, n = 4

f(4) = (4 - 1) + f(4 - 1) = 3 + f(3) = 3 + 3 = 6

Similarly, f(5) = 10, f(6) = 15, f(7) = 21, f(8) = 28

Therefore, above pattern can be written in the form of

\(f(3) = \frac{3\;(3\;-\;1)}{2} = 3\)

\(f(4) = \frac{4\;(4\;-\;1)}{2} = 6\)

\(f(5) = \frac{5\;(5\;-\;1)}{2} = 10\)

In general

\(f(n) = \frac{n\;(n\;-\;1)}{2} \)