Ask question

Ask question

All categories

3 years ago

9

Which sequence is generated by the function f(n + 1) = f(n) – 2 for f(1) = 10?

1 answer:

kozerog [31]3 years ago

7 0

F(n+1)=f(n)-2 is the recursive function for this arithmetic sequence. The explicit form is:

f(n)=a-d(n-1) where a(initial term) is 10 and d(common difference) is -2

f(n)=10-2(n-1) which cleans up to

f(n)=12-2n so the first few terms are:

10,8,6,4,2,0,-2,-4.....

You might be interested in

What is the slope of the line shown below?

erik [133]

Answer:

C) 2

Step-by-step explanation:

7 0

3 years ago

Pls help and show workings due ASAP

Fed [463]

Answer:

goodluck

Step-by-step explanation:

7 0

2 years ago

What is the probability of getting two heads and two tails in four flips of a fair coin?

jekas [21]

Number of combinations possible = 4! / 2! 2! = 6

Each of these has probability 1/2*1/2*1/2*1/2 = 1/16

Required probability = 6 * 1/16 = 3/8

3 0

3 years ago

HELP PLS I NEED THIS DONE ASAP

Sholpan [36]

Answer:

a = 4, b = 0

Also, a = 0 and b = 0 :)

8 0

3 years ago

The sequence$$1,2,1,2,2,1,2,2,2,1,2,2,2,2,1,2,2,2,2,2,1,2,\dots$$consists of $1$'s separated by blocks of $2$'s with $n$ $2$'s i

kicyunya [14]

Consider the lengths of consecutive 1-2 blocks.

block 1 - 1, 2 - length 2

block 2 - 1, 2, 2 - length 3

block 3 - 1, 2, 2, 2 - length 4

block 4 - 1, 2, 2, 2, 2 - length 5

and so on.

Recall the formula for the sum of consecutive positive integers,

$\displaystyle \sum_{i=1}^j i = 1 + 2 + 3 + \cdots + j = \frac{j(j+1)}2 \implies \sum_{i=2}^j = \frac{j(j+1) - 2}2$

Now,

$1234 = \dfrac{j(j+1)-2}2 \implies 2470 = j(j+1) \implies j\approx49.2016$

which means that the 1234th term in the sequence occurs somewhere about 1/5 of the way through the 49th 1-2 block.

In the first 48 blocks, the sequence contains 48 copies of 1 and 1 + 2 + 3 + ... + 47 copies of 2, hence they make up a total of

$\displaystyle \sum_{i=1}^48 1 + \sum_{i=1}^{48} i = 48+\frac{48(48+1)}2 = 1224$

numbers, and their sum is

$\displaystyle \sum_{i=1}^{48} 1 + \sum_{i=1}^{48} 2i = 48 + 48(48+1) = 48\times50 = 2400$

This leaves us with the contribution of the first 10 terms in the 49th block, which consist of one 1 and nine 2s with a sum of $1+9\times2=19$ .

So, the sum of the first 1234 terms in the sequence is 2419.

8 0

2 years ago