a) use the binomial formula to find the 5th term of the expansion (2x - 3)^8 ------ i got 90720x?

b) Find a_n, the general term for a geometric sequence, given that the fourth term, a_4 is 16 and the seventh term is -128

c) Find the sum of 8+12+16+20+24+.....to 15 terms

thanks!

Almost: C(8,4)(2x)^4(-3)^4 90720x^4

ar^3 = 16
ar^6 = -128
r^3 = -8
r = -2
so, a = -2

a_n = -2*(-2)^(n-1) = -(-2)^n

S15 = 15/2 (2*8+14*4) = 540

D expression for (2x-3)^8 =(X)^8-8(X)^7(Y) 28(X)^6(Y)^2....70(X)^3(Y)^5... Put X=2x nd y=-3...there4 d 5th term =70(X)^3(Y)^5...=70*8x^3*-243=-136080..it seems the anxa z -136080.. Read more on pascal triangle or vandalmodes theorem

a) To find the 5th term of the expansion (2x - 3)^8 using the binomial formula, you need to use the formula:

Term = (nCr) * (a^r) * (b^(n-r))

Where:
- n is the exponent of the binomial, in this case, 8.
- r is the term number you want to find, in this case, 5.
- nCr is the number of ways to choose r items from a set of n items.
- a is the coefficient of the first term, in this case, 2x.
- b is the coefficient of the second term, in this case, -3.

Using this formula, the 5th term is:
Term = (8C5) * (2x)^5 * (-3)^3

Simplifying this expression, we get:
Term = (8! / (5! * 3!)) * (32x^5) * (-27)

Simplifying the factorial, we have:
Term = (8 * 7 * 6) / (3 * 2 * 1) * (-27) * 32x^5

Simplifying further, we get:
Term = 56 * (-27) * 32x^5

Calculating, the 5th term of the expansion is:
Term = -48384x^5

So, the correct answer is -48384x^5.

b) To find the general term (a_n) of a geometric sequence given the fourth term (a_4) as 16 and the seventh term as -128, we need to use the formula for the nth term of a geometric sequence:

a_n = a_1 * r^(n-1)

In this formula:
- a_n is the general term.
- a_1 is the first term.
- r is the common ratio.
- n is the term number.

Given that a_4 is 16 and a_7 is -128, we can write the following equations:
16 = a_1 * r^(4-1)
-128 = a_1 * r^(7-1)

Now, we can divide the second equation by the first equation to find the common ratio, r:

-128 / 16 = (a_1 * r^(7-1)) / (a_1 * r^(4-1))
-8 = r^3

Taking the cube root of both sides, we get:
-2 = r

Now, we can substitute the common ratio back into the first equation to solve for a_1:

16 = a_1 * (-2)^(4-1)
16 = a_1 * (-2)^3
16 = a_1 * (-8)
a_1 = -2

Therefore, the general term (a_n) of the geometric sequence is:
a_n = -2 * (-2)^(n-1)

c) To find the sum of the arithmetic sequence 8, 12, 16, 20, 24, ..., up to 15 terms, we can use the formula for the sum of an arithmetic series:

Sum = (n/2) * (first term + last term)

In this case, the first term is 8 and the last term is obtained by plugging in the given number of terms (15) into the arithmetic sequence formula:

Last term = first term + (n-1) * common difference

The common difference is obtained by subtracting the first term from the second term in the given sequence:

Common difference = 12 - 8 = 4

Plugging these values into the formulas, we have:

Last term = 8 + (15-1) * 4
Last term = 8 + 14 * 4
Last term = 8 + 56
Last term = 64

Now, we can calculate the sum:

Sum = (15/2) * (8 + 64)
Sum = 7.5 * 72
Sum = 540

Therefore, the sum of the arithmetic sequence 8, 12, 16, 20, 24, ..., up to 15 terms is 540.

a) To find the 5th term of the expansion of (2x - 3)^8 using the binomial formula, we can use the formula:

T(n+1) = C(n, r) * a^(n-r) * b^r

where T(n+1) is the (n+1)th term, C(n, r) is the binomial coefficient, a is the coefficient of the first term, b is the coefficient of the second term, n is the power, and r is the term number.

In this case, the term number is 5, so we substitute n = 8, r = 5, a = 2x, and b = -3 into the formula:

T(5+1) = C(8, 5) * (2x)^(8-5) * (-3)^5

Simplifying further:

T(6) = C(8, 5) * (2x)^3 * (-3)^5

C(8, 5) is the binomial coefficient and can be calculated as:

C(8, 5) = 8! / (5! * (8-5)!) = 8! / (5! * 3!)

Plug in the values:

T(6) = 56 * 8 * x^3 * (-3)^5

Simplifying even further:

T(6) = 56 * 8 * x^3 * (-243)

T(6) = -10368x^3

Therefore, the 5th term of the expansion (2x - 3)^8 is -10368x^3.

b) To find the general term a_n for a geometric sequence, we need at least two terms. Given that the fourth term (a_4) is 16 and the seventh term is -128, we can use these two terms to find the common ratio (r).

The general formula for a geometric sequence is given by:

a_n = a_1 * r^(n-1)

where a_n is the nth term, a_1 is the first term, r is the common ratio, and n is the term number.

Let's use the fourth term to find the common ratio:

a_4 = a_1 * r^(4-1)
16 = a_1 * r^3

Similarly, using the seventh term:

a_7 = a_1 * r^(7-1)
-128 = a_1 * r^6

Now we have a system of two equations:

16 = a_1 * r^3
-128 = a_1 * r^6

Dividing the second equation by the first equation:

(-128) / 16 = (a_1 * r^6) / (a_1 * r^3)
-8 = r^3

Taking the cube root of both sides:

r = -2

Now, substitute the value of r into one of the original equations:

16 = a_1 * (-2)^3
16 = a_1 * (-8)
16 = -8a_1

Solving for a_1:

a_1 = 16 / -8
a_1 = -2

So, the first term (a_1) of the sequence is -2 and the common ratio (r) is -2.

The general term a_n can now be written as:

a_n = -2 * (-2)^(n-1)

Therefore, the general term for the geometric sequence is a_n = -2 * (-2)^(n-1).

c) To find the sum of the arithmetic sequence 8, 12, 16, 20, 24, ..., up to 15 terms, we can use the formula for the sum of an arithmetic series:

Sn = (n/2) * (a_1 + a_n),

where Sn is the sum of the series, n is the number of terms, a_1 is the first term, and a_n is the nth term.

In this case, we have:

n = 15, a_1 = 8, and a_n = 8 + (n-1)d, where d is the common difference.

The common difference can be found by subtracting the second term from the first:

d = 12 - 8 = 4.

Now, we can find a_n:

a_n = 8 + (n-1)d
a_n = 8 + (15-1)4
a_n = 8 + 14 * 4
a_n = 8 + 56
a_n = 64.

Now we have all the values needed to calculate the sum:

Sn = (n/2) * (a_1 + a_n)
Sn = (15/2) * (8 + 64)
Sn = (15/2) * 72
Sn = 540.

Therefore, the sum of the arithmetic series 8, 12, 16, 20, 24, ..., up to 15 terms is 540.