Question

Consider the following sequence

Answer 1

9514 1404 393

Answer:

  2n/2^n

Step-by-step explanation:

It appears that the terms are ...

  $\dfrac{1}{1},\ \dfrac{2}{2},\ \dfrac{3}{4},\ \dfrac{4}{8},\ \dots$

That is, the numerator is counting numbers, and the denominator is powers of 2.

  $f[n]=\dfrac{n}{2^{(n-1)}}\quad\text{or}\quad f[n]=2n\cdot2^{-n}$

Answer 2

SEQUENCE:

• AP —» 1/1 , 2/2, 3/4, 4/8...

ANSWER:

We are given with a sequence, so we have to find the first term and common difference.

• a = 1/1.
• d = 2/2 - 1/1 = 1 - 2 /2 = - 1/2.

So, Formula to finding nth term

Tn = a + (n - 1)d

4/8 = 1/1 + (n - 1) - 1/2

4/8 = 1/1 - n/2 + 1/2

4/8 = 2 + 1 /2 - n/2

4/8 = 3/2 - n/2

3/2 - 4/8 = n/2

12 - 4/8 = n/2

8/8 = n/2

1 = n/2

n = 2.