Mathematics: Find the 8th term in the expansion (x-2)^9

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07.02.2020

Find the 8th term in the expansion (x-2)^9

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Mathematics

Find the 8th term in the expansion (x + 3y)2

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P Answered by Specialist

Find the 8th term of the expansion of the binomial (2x - y) 14 Find the third term of expansion of the binomial (x + 3y) 10 Back to the Sequences and Series Home Page. Back to the Intermediate Algebra (Math 154)

Step-by-step explanation:

Mathematics

Find the 8th term of the expansion a. 5x3y7 b. -14x3y7 c. 70x5y5 d. -15x3y7 select the best answer from the choices provided a b c d

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P Answered by PhD

Correct choice is D

Step-by-step explanation:

The i-th term of the binomial expansion $\left(\dfrac{1}{2}x-y\right)^{10}$ is

$T_i=C^{10}_{i-1}\cdot \left(\dfrac{1}{2}x\right)^{10+1-i}\cdot (-y)^{i-1}.$

If i=8, then

$T_i=C^{10}_{7}\cdot \left(\dfrac{1}{2}x\right)^{10+1-8}\cdot (-y)^{8-1}=-120\cdot \left(\dfrac{1}{2}x\right)^3\cdot y^7=-120\cdot \dfrac{1}{8}x^3\cdot y^7=-15x^3y^7.$

Mathematics

Find the 8th term of the expansion of (1/2x-y)^10 a 5x^3y^7 b 70x5y5 c -14x3y7 d -15x3y7

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P Answered by Master

d. -15x^3y^7

Step-by-step explanation:

By the binomial theorem,

$(p+q)^n=\sum_{r=0}^{n} ^nC_r (p)^{n-r} q^r$

Where,

$^nC_r=\frac{n!}{r!(n-r)!}$

Thus,

$(\frac{1}{2}x-y)^{10}=(\frac{x}{2}+(-y))^{10})=\sum_{r=0}^{n} ^10C_r (\frac{x}{2})^{10-r} (-y)^{r}$

For the 8th term, r = 7,

Thus, the 8th term would be,

$^{10}C_7 (\frac{x}{2})^{10-7} (-y)^{7}$

$=\frac{10!}{7!(10-7)!} (\frac{x}{2})^3 (-y)^7$

$=-\frac{10\times 9\times 8}{3\times 2\times 1}\times \frac{x^3}{8}\times y^7$

$=-\frac{720x^3y^7}{48}$

=-15x^3y^7

Hence, option 'D' is correct.

Mathematics

Find the 8th term of the expansion a. 5x3y7 b. -14x3y7 c. 70x5y5 d. -15x3y7 select the best answer from the choices provided a b c d

Step-by-step answer

P Answered by PhD

Correct choice is D

Step-by-step explanation:

The i-th term of the binomial expansion $\left(\dfrac{1}{2}x-y\right)^{10}$ is

$T_i=C^{10}_{i-1}\cdot \left(\dfrac{1}{2}x\right)^{10+1-i}\cdot (-y)^{i-1}.$

If i=8, then

$T_i=C^{10}_{7}\cdot \left(\dfrac{1}{2}x\right)^{10+1-8}\cdot (-y)^{8-1}=-120\cdot \left(\dfrac{1}{2}x\right)^3\cdot y^7=-120\cdot \dfrac{1}{8}x^3\cdot y^7=-15x^3y^7.$

Mathematics

Find the 8th term of the expansion of (1/2x-y)^10 a 5x^3y^7 b 70x5y5 c -14x3y7 d -15x3y7

Step-by-step answer

P Answered by Specialist

d. -15x^3y^7

Step-by-step explanation:

By the binomial theorem,

$(p+q)^n=\sum_{r=0}^{n} ^nC_r (p)^{n-r} q^r$

Where,

$^nC_r=\frac{n!}{r!(n-r)!}$

Thus,

$(\frac{1}{2}x-y)^{10}=(\frac{x}{2}+(-y))^{10})=\sum_{r=0}^{n} ^10C_r (\frac{x}{2})^{10-r} (-y)^{r}$

For the 8th term, r = 7,

Thus, the 8th term would be,

$^{10}C_7 (\frac{x}{2})^{10-7} (-y)^{7}$

$=\frac{10!}{7!(10-7)!} (\frac{x}{2})^3 (-y)^7$

$=-\frac{10\times 9\times 8}{3\times 2\times 1}\times \frac{x^3}{8}\times y^7$

$=-\frac{720x^3y^7}{48}$

=-15x^3y^7

Hence, option 'D' is correct.

Mathematics

Find a if 17th and 18th terms in the expansion of (2+a)^50 are equal.

Step-by-step answer

P Answered by PhD

The 17th term is 50C16*2^34*a^16
The 18th term is 50C17*2^33*a^17

The ratio of these two terms is (17th term)/(18th term) = 1/a. We want the value of this to be 1, so
.. A = 1

My calculator says
.. 50C16 = 4,923,689,695,575
.. 50C17 = 9,847,379,391,150
the latter being exactly 2 times the former.

Mathematics

Find a if 17th and 18th terms in the expansion of (2+a)^50 are equal.

Step-by-step answer

P Answered by PhD

Mathematics

1)Find the seventh term in the expansion (3x - (2y)/(sqrt(3) * x)) ^ 14 Simplify your answer. 2) In the expansion of (sqrt(x) + 3/(sqrt(x))) ^ 8 , determine the coefficient of x ^ 2 3) Solve the following inequality: (x + 4)/(4x ^ 2 - 4x + 1) smaller than or equal - 2/((2x - 1) ^ 2 * (x + 1)) 4) The fifth term and the fifteen term of an arithmetic sequence are -54 and 36 respectively (i) find the first term and the common difference of the sequence. (ii)Determine the twentieth term of the sequence. (iii)Determine the sum from 8th term until the

Step-by-step answer

P Answered by PhD

r+1=7; r=6.
(11/7).(-3x)^11-6.(-2y)^6;
11.10.9.8.7/7(cut by seven).(-3x)^5.(-2y)^6;
7920.-3^5.x^5.2^6.y^6.

Mathematics

Find the term indicated in the expansion. (x - 3y)11; 8th term

Step-by-step answer

P Answered by Specialist

The term in the expansion:
T ( k+1) = n C k * A^(n-k) * B^k.
In this case: n = 11, k + 1 = 8, so k = 7.
A = x, B = - 3 y
T 8 = 11 C 7 * x^(11-7) * ( - 3 y )^7 =
=( 11 *10 * 9 * 8 * 7 * 6 * 5 ) / ( 7 * 6 * 5 * 4 * 3 * 2 * 1 )* x^4 * ( - 2,187 y^7 ) =
= 330 * ( - 2,187 ) x^4 y^7 = - 721,710 x^4 y^7
The 8th term in expansion is

Mathematics

What is the 8th term of the binomial expansion (x + y)10?

Step-by-step answer

P Answered by Specialist

120x^3y^7

We can use the Pascal's triangle to solve this question.

This pascal's triangle is shown in the Image below. To build the triangle, begin with the number 1 at the top, then continue placing numbers below it in a triangular pattern. In this way, each number are the numbers directly above added together. So, the expression is :

(x+y)10

After you do the steps you be left with:

120x^3y^7