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3 years ago

Please help with this! I will mark you as brainliest!

Mathematics

2 answers:

Svetllana [295]3 years ago

3 0

Answer:

an = 5 + (n-1)7

= 5+7n-7

= 7n-2

puteri [66]3 years ago

3 0

Answer:

$a_{n}$ = 7n - 2

Step-by-step explanation:

The n th term of an arithmetic sequence is

$a_{n}$ = a₁ + (n - 1)d

where a₁ is the first term and d the common difference

Here a₁ = 5 and d = 12 - 5 = 7, thus

$a_{n}$ = 5 + 7(n - 1) = 5 + 7n - 7 = 7n - 2

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Binomial Expansion/Pascal's triangle. Please help with all of number 5.

Mandarinka [93]

$\begin{matrix}1\\1&1\\1&2&1\\1&3&3&1\\1&4&6&4&1\end{bmatrix}$

The rows add up to $1,2,4,8,16$ , respectively. (Notice they're all powers of 2)

The sum of the numbers in row $n$ is $2^{n-1}$ .

The last problem can be solved with the binomial theorem, but I'll assume you don't take that for granted. You can prove this claim by induction. When $n=1$ ,

$(1+x)^1=1+x=\dbinom10+\dbinom11x$

so the base case holds. Assume the claim holds for $n=k$ , so that

$(1+x)^k=\dbinom k0+\dbinom k1x+\cdots+\dbinom k{k-1}x^{k-1}+\dbinom kkx^k$

Use this to show that it holds for $n=k+1$ .

$(1+x)^{k+1}=(1+x)(1+x)^k$
$(1+x)^{k+1}=(1+x)\left(\dbinom k0+\dbinom k1x+\cdots+\dbinom k{k-1}x^{k-1}+\dbinom kkx^k\right)$
$(1+x)^{k+1}=1+\left(\dbinom k0+\dbinom k1\right)x+\left(\dbinom k1+\dbinom k2\right)x^2+\cdots+\left(\dbinom k{k-2}+\dbinom k{k-1}\right)x^{k-1}+\left(\dbinom k{k-1}+\dbinom kk\right)x^k+x^{k+1}$

Notice that

$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!}{\ell!(k-\ell)!}+\dfrac{k!}{(\ell+1)!(k-\ell-1)!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(\ell+1)}{(\ell+1)!(k-\ell)!}+\dfrac{k!(k-\ell)}{(\ell+1)!(k-\ell)!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(\ell+1)+k!(k-\ell)}{(\ell+1)!(k-\ell)!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(k+1)}{(\ell+1)!(k-\ell)!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{(k+1)!}{(\ell+1)!((k+1)-(\ell+1))!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dbinom{k+1}{\ell+1}$

So you can write the expansion for $n=k+1$ as

$(1+x)^{k+1}=1+\dbinom{k+1}1x+\dbinom{k+1}2x^2+\cdots+\dbinom{k+1}{k-1}x^{k-1}+\dbinom{k+1}kx^k+x^{k+1}$

and since $\dbinom{k+1}0=\dbinom{k+1}{k+1}=1$ , you have

$(1+x)^{k+1}=\dbinom{k+1}0+\dbinom{k+1}1x+\cdots+\dbinom{k+1}kx^k+\dbinom{k+1}{k+1}x^{k+1}$

and so the claim holds for $n=k+1$ , thus proving the claim overall that

$(1+x)^n=\dbinom n0+\dbinom n1x+\cdots+\dbinom n{n-1}x^{n-1}+\dbinom nnx^n$

Setting $x=1$ gives

$(1+1)^n=\dbinom n0+\dbinom n1+\cdots+\dbinom n{n-1}+\dbinom nn=2^n$

which agrees with the result obtained for part (c).

4 0

3 years ago

Is algebra.

Nostrana [21]

Answer:

1) b 2) b

Step-by-step explanation:

1) Both expressions have (x+6). Rearrange them and you'll have one expression as (x+6) and the other as (5ab-4).

2) (4b - 7x)(a + b) factors to be 4ba + 8b - 7ax - 14x, which can be rearranged to 4ab - 7ax + 8b - 14x

3 0

2 years ago

Read 2 more answers

Kamal invested $1200 in a savings acc paying 1.8% per yr compound interest.

Greeley [361]

Answer:

A = $1311 and 96cents

Step-by-step explanation:

$A = P(1 + \frac{r}{n})^{nt}$

P = $1200

r = 1.8% = 0.018

n = 1 (compounded yearly)

t = 5

$A = 1200(1 + 0.018)^5 = 1200 \times(1.018)^5 = 1311. 96$

7 0

3 years ago

7y-3x= -5 solve for x

nika2105 [10]

7y - 3x = -5
-3x = -7y - 5
x = 7/3y + 5/3

7 0

3 years ago

Solve:<br> -11 - 79 +5g = 3

Katyanochek1 [597]

Answer:

g=18.6

Step-by-step explanation:

-11-79+5g=3

-90+5g=3

5g=3+90

5g=93

g=93/5=18.6

g=18.6

6 0

3 years ago