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astra-53 [7]
3 years ago
10

Find the coefficient of x^5y^8 in (x+y)^13

Mathematics
2 answers:
motikmotik3 years ago
7 0

Binomial Theorem:

(x+y)^n\quad=\quad\sum\limits_k^n\left(\begin{array}{cc}n\\k\end{array}\right) x^{n-k} y^k


Where the coefficient is defined to be:

\left(\begin{array}{cc}n\\k\end{array}\right)\quad=\quad \dfrac{n!}{k!(n-k)!}


Notice that the exponent on y is 8 for the term they gave us.

This is the term which corresponds to k=8.

\left(\begin{array}{cc}13\\8\end{array}\right)x^{13-8}y^8


So our coefficient is:

\left(\begin{array}{cc}13\\8\end{array}\right)\quad=\quad\dfrac{13!}{8!(13-8)!}


If you expand the factorial in the numerator,

you can cancel some stuff out,

\dfrac{13\cdot12\cdot11\cdot10\cdot9\cdot8!}{8!\cdot5!}\quad=\quad \dfrac{13\cdot12\cdot11\cdot10\cdot9}{5!}


Further simplification should get you to this:

13\cdot11\cdot3\cdot3\quad=\quad1287

Dafna1 [17]3 years ago
5 0

recall that the exponent for the first term will be starting at 13, and descending by one on every expanded term, in the binomial theorem.


so, 13, 12, 11, 10, 9, 8, 7, 6, 5 <---- x⁵ will then be in the 9th term


now, using the "combination" formula to get the coefficient of the 9th term.


k = 8 <--- the 9th term


n = 13 <--- highest exponent


\bf \textit{the coefficient and values of an expanded term} \\\\ (x+y)^{13} \qquad \qquad  \begin{array}{llll} expansion\\ for\\ 9^{th}~term \end{array} \quad  \begin{cases} \stackrel{term}{k}=0..n\\ \stackrel{exponent}{n}\\ -----\\ k=\stackrel{9~term}{8}\\ n=13 \end{cases}


\bf \stackrel{coefficient}{\left(\frac{n!}{k!(n-k)!}\right)} \qquad \stackrel{\stackrel{first~term}{factor}}{\left( a^{n-k} \right)} \qquad \stackrel{\stackrel{second~term}{factor}}{\left( b^k \right)} \\\\\\ \cfrac{13!}{8!(13-8)!}~~x^{13-8}~~(+y)^8\implies 1287~x^5y^8

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QUESTION IN THE ATTACHMENT
eimsori [14]

Answer:

A. The sum of the first 10th term is 100.

B. The sum of the nth term is n²

Step-by-step explanation:

Data obtained from the question include:

Sum of 20th term (S20) = 400

Sum of 40th term (S40) = 1600

Sum of 10th term (S10) =..?

Sum of nth term (Sn) =..?

Recall:

Sn = n/2[2a + (n – 1)d]

Sn is the sum of the nth term.

n is the number of term.

a is the first term.

d is the common difference

We'll begin by calculating the first term and the common difference. This is illustrated below:

Sn = n/2 [2a + (n – 1)d]

S20 = 20/2 [2a + (20 – 1)d]

S20= 10 [2a + 19d]

S20 = 20a + 190d

But:

S20 = 400

400 = 20a + 190d .......(1)

S40 = 40/2 [2a + (40 – 1)d]

S40 = 20 [2a + 39d]

S40 = 40a + 780d

But

S40 = 1600

1600 = 40a + 780d....... (2)

400 = 20a + 190d .......(1)

1600 = 40a + 780d....... (2)

Solve by elimination method

Multiply equation 1 by 40 and multiply equation 2 by 20 as shown below:

40 x equation 1:

40 x (400 = 20a + 190d)

16000 = 800a + 7600. ........ (3)

20 x equation 2:

20 x (1600 = 40a + 780d)

32000 = 800a + 15600d......... (4)

Subtract equation 3 from equation 4

Equation 4 – Equation 3

32000 = 800a + 15600d

– 16000 = 800a + 7600d

16000 = 8000d

Divide both side by 8000

d = 16000/8000

d = 2

Substituting the value of d into equation 1

400 = 20a + 190d

d = 2

400 = 20a + (190 x 2)

400 = 20a + 380

Collect like terms

400 – 380 = 20a

20 = 20a

Divide both side by 20

a = 20/20

a = 1

Therefore,

First term (a) = 1.

Common difference (d) = 2.

A. Determination of the sum of the 10th term.

First term (a) = 1.

Common difference (d) = 2

Number of term (n) = 10

Sum of 10th term (S10) =..?

Sn = n/2 [2a + (n – 1)d]

S10 = 10/2 [2x1 + (10 – 1)2]

S10 = 5 [2 + 9x2]

S10 = 5 [2 + 18]

S10 = 5 x 20

S10 = 100

Therefore, the sum of the first 10th term is 100.

B. Determination of the sum of the nth term.

First term (a) = 1.

Common difference (d) = 2

Sum of nth term (Sn) =..?

Sn = n/2 [2a + (n – 1)d]

Sn = n/2 [2x1 + (n – 1)2]

Sn = n/2 [2 + 2n – 2]

Sn = n/2 [2 – 2 + 2n ]

Sn = n/2 [ 2n ]

Sn = n²

Therefore, the sum of the nth term is n²

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Answer:

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