Alpha writes the infinite arithmetic sequence 10, 8, 6, 4, 2, 0...

Beta writes the infinite geometric sequence 9, 6, 4, 8/3, 16/9,...

Gamma makes a sequence whose n^th term is the product of the n^th term of Alpha's sequence and the n^th term of Beta's sequence: 10*9,8*6,6*4,4*8/3,2*16/9,...

What is the sum of Gamma's entire sequence?

I know this is a really long question but I need help.

Starting with n=0,

An = 10-2n
Bn = 9*(2/3)^n
Gn = 18(5-n)*(2/3)^n

By the ratio test, the series converges.
Sum = 162

No problem! Let's break it down step-by-step.

To find the sum of Gamma's entire sequence, we need to find the sum of each term in the sequence.

The n-th term of Gamma's sequence is given by multiplying the n-th term of Alpha's sequence with the n-th term of Beta's sequence.

First, let's write down the sequences:

Alpha: 10, 8, 6, 4, 2, 0, ...
Beta: 9, 6, 4, 8/3, 16/9, ...

Now, let's find the first few terms in Gamma's sequence by multiplying the corresponding terms from Alpha and Beta:

Gamma: (10 * 9), (8 * 6), (6 * 4), (4 * 8/3), (2 * 16/9), ...

Simplifying, we get:
Gamma: 90, 48, 24, 32/3, 32/9, ...

To find the sum of this sequence, we can rewrite it as a sum of fractions:

Gamma: 90/1, 48/1, 24/1, 32/3, 32/9, ...

Next, let's find a common denominator for all the fractions:

Gamma: (90/1 * 9/9), (48/1 * 1/1), (24/1 * 3/3), (32/3 * 3/3), (32/9 * 1/1), ...

Simplifying, we get:
Gamma: 810/9, 48/1, 72/3, 96/9, 32/9, ...

Now, let's add up all the terms in the sequence:
Gamma: 810/9 + 48/1 + 72/3 + 96/9 + 32/9, ...

To find the sum, we need to find a common denominator for all the fractions:

Gamma: (810/9 * 1/1), (48/1 * 9/9), (72/3 * 3/3), (96/9 * 1/1), (32/9 * 3/3), ...

Simplifying, we get:
Gamma: 810/9, 432/9, 216/9, 96/9, 96/9, ...

Now, let's add up all the fractions:

Gamma: (810 + 432 + 216 + 96 + 96)/9, ...

Simplifying, we get:
Gamma: 1650/9, ...

To find the final sum, we divide the numerator by the denominator:

Gamma: 1650/9 = 183.33...

So, the sum of Gamma's entire sequence is approximately 183.33...

To find the sum of Gamma's entire sequence, we need to find a pattern in the sequence and use a formula to calculate the sum.

Let's first write out the terms of Gamma's sequence:

Gamma's sequence: (10*9), (8*6), (6*4), (4*8/3), (2*16/9), ...

We can simplify each term:

Gamma's sequence: 90, 48, 24, 32/3, 32/9, ...

Now, let's analyze the pattern in Gamma's sequence. Notice that each term is obtained by multiplying the corresponding terms of Alpha's and Beta's sequences. We can write the general formula for the n-th term of Gamma's sequence as:

Gamma(n) = Alpha(n) * Beta(n)

For example, the n-th term of Gamma's sequence is given by:

Gamma(n) = 10 * 9 * 8 * 6 * ... * (10 - (n - 1)*2) * ... * (9/3 + (n - 1)*4/3) * ... * (16/9 + (n - 1)*8/9)

Now, let's find the sum of Gamma's entire sequence. We can use the formula for the sum of an infinite geometric series:

sum = a / (1 - r)

where a is the first term of the series and r is the common ratio.

In our case, the first term (a) of Gamma's sequence is 90. To find the common ratio (r), we can divide the n-th term by the (n-1)-th term:

r = (48 / 90) = (24 / 48) = (32 / 24) = (32/9) / (24/9) = (32/9) * (9/24) = 32 / 24 = 4/3

Now we can substitute these values into the formula to find the sum of Gamma's sequence:

sum = 90 / (1 - (4/3))

To simplify this expression, we need to find a common denominator:

sum = 90 / (3/3 - 4/3)
= 90 / (-1/3)
= -270

Therefore, the sum of Gamma's entire sequence is -270.

I really don't know the answer?