NCERT Class XI Mathematics - Sequences and Series - Solutions
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Question : 85
Total: 106
If f is a function satisfying f (x + y) = f (x) f (y) for all x, y ∈ N such that f (1) = 3 and
f ( x ) = 120, find the value of n.
Solution:
We are given
f (x + y) = f (x) f (y) ∀ x, y ∈ N ....(i)
f (1) = 3 ,
f (x) = 120
∴ f (2) = f (1 + 1) = f (1) f (1) = 3·3 =3 2 [By using (i)]
f (3) = f (2 + 1) = f (2)· f (1) =3 2 · 3 = 3 3
Proceeding like above, we get f (n) =3 n
We have
f (x) = 120
⇒ f (1) + f (2) + ...... + f (n) = 120 ⇒ 3 +3 2 + ...... + 3 n = 120
⇒
= 120 ⇒ 3 ( 3 n − 1 ) = 120 × 2
⇒3 n − 1 = 80 ⇒ 3 n = 81 ⇒ 3 n = ( 3 ) 4
∴ n = 4
f (x + y) = f (x) f (y) ∀ x, y ∈ N ....(i)
f (1) = 3 ,
∴ f (2) = f (1 + 1) = f (1) f (1) = 3·3 =
f (3) = f (2 + 1) = f (2)· f (1) =
Proceeding like above, we get f (n) =
We have
⇒ f (1) + f (2) + ...... + f (n) = 120 ⇒ 3 +
⇒
⇒
∴ n = 4
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