Question

If f (0) = 1, f (1) = 4 and f (x) = f(x - 2) + 6 for all integers x > 1, then f(71)

= ?​

Answer 1

Answer:

f(2)=f(0) +6. for x=2

f(2)=7

f(3)=f(1)+6=4+6=10

then similary

f(4)=13 and f(5)=16

here

7,10,13,16.......these are making Ap series

then

formula of nth term

a+(n-1)d=An

7+(71-1)3=A71

f(71)=217

I hope u get understand

Answer 2

The value of f(71)=211

Explanation:

Given : f (0) = 1, f (1) = 4 and $f (x) = f(x-2) + 6$

At x= 2 ,

$f (2) = f(0) + 6=1+6=7$

At x= 3 ,

$f (2) = f(1) + 6= 4+6=10$

At x= 4 ,

$f (2) = f(2) + 6= 7+6=13$

The sequence becomes 1 , 4, 7 , 10 , 13 , .... so on.

We can see there the difference between the terms are same as 3.

Therefore it is a arithmetic sequence.

The nth term in AP= $a_n= a+(n-1)d$ , where a= First term , d= common difference

For 71 th term ,  $a_{71}= 1+(71-1)(3)=211$

Therefore , The value of  f(71)=211 .

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If f(x)=x^2 - 4x +6, find f(1) - f(-1)

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