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pogonyaev
3 years ago
10

Can you guys help we with this question. ​​​​​​

Mathematics
2 answers:
Natasha2012 [34]3 years ago
6 0
4r-2n=2

4r-10=2

4r=12

R=13
satela [25.4K]3 years ago
4 0

Answer:

40

Step-by-step explanation:

(2x+1/(2x))^5 *(2x -1/(2x))^5

= ((2x)^2 -1/(2x)^2)^5 (a+b)*(a-b) =a2-b2

= (4x^2-1/4(x)^2)^5

now

x =4x^2. ,a = 1/4(x)^2 ,n =5

we have

general term = Cr *x^r *a^(n-r)

= Cr * (4x^2)^r * (1/4(x)^2)^(n-r)

= Cr *4^r * X^2r * 1/( 4^(n-r) *x^(2n-2r)

= Cr * 4^r/4^(n-r) * x^(2r)/x^(2n-2r)

= Cr * 4(2r-n) *x(4r-2n)

now for x^2

4r-2n = 2

4r -10=2

4r =12

r = 3

now for coeff

C(5,3) * 4^(2*3-5)

5!/(3!*(5-3)!) * 4

5*4/(2*1)*4

40

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Which recursive sequence would produce the sequence 8,-35,137,…?
vagabundo [1.1K]

The recursive sequence that would produce the sequence 8,-35,137,… is T(n + 1) = -3 - 4T(n) where T(1) = 8

<h3>How to determine the recursive sequence that would produce the sequence?</h3>

The sequence is given as:

8,-35,137,…

From the above sequence, we can see that:

The next term is the product of the current term and -4 added to -3

i.e.

Next term = -3 + Current term * -4

So, we have:

T(n + 1) = -3 + T(n) * -4

Rewrite as:

T(n + 1) = -3 - 4T(n)

Hence, the recursive sequence that would produce the sequence 8,-35,137,… is T(n + 1) = -3 - 4T(n) where T(1) = 8

Read more about recursive sequence at

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Step-by-step explanation:

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