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3 years ago

Use the numbers below to simplify the expression. Numbers may be used once or not at all. 34671011 3y + 7 + 4y + 3 = y +

Mathematics

1 answer:

Svetllana [295]3 years ago

8 0

Answer:

$3y + 7 + 4y + 3 = 7y + 10$

Step-by-step explanation:

Given

$3y + 7 + 4y + 3 = [\ ]y + [\ ]$

Required

Simplify

We have:

$3y + 7 + 4y + 3$

Collect like terms

$3y + 7 + 4y + 3 = 3y + 4y + 7 + 3$

Evaluate like terms

$3y + 7 + 4y + 3 = 7y + 10$

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3 years ago

If S_1=1,S_2=8 and S_n=S_n-1+2S_n-2 whenever n≥2. Show that S_n=3⋅2n−1+2(−1)n for all n≥1.

Snezhnost [94]

You can try to show this by induction:

• According to the given closed form, we have $S_1=3\times2^{1-1}+2(-1)^1=3-2=1$ , which agrees with the initial value S₁ = 1.

• Assume the closed form is correct for all n up to n = k. In particular, we assume

$S_{k-1}=3\times2^{(k-1)-1}+2(-1)^{k-1}=3\times2^{k-2}+2(-1)^{k-1}$

and

$S_k=3\times2^{k-1}+2(-1)^k$

We want to then use this assumption to show the closed form is correct for n = k + 1, or

$S_{k+1}=3\times2^{(k+1)-1}+2(-1)^{k+1}=3\times2^k+2(-1)^{k+1}$

From the given recurrence, we know

$S_{k+1}=S_k+2S_{k-1}$

so that

$S_{k+1}=3\times2^{k-1}+2(-1)^k + 2\left(3\times2^{k-2}+2(-1)^{k-1}\right)$

$S_{k+1}=3\times2^{k-1}+2(-1)^k + 3\times2^{k-1}+4(-1)^{k-1}$

$S_{k+1}=2\times3\times2^{k-1}+(-1)^k\left(2+4(-1)^{-1}\right)$

$S_{k+1}=3\times2^k-2(-1)^k$

$S_{k+1}=3\times2^k+2(-1)(-1)^k$

$\boxed{S_{k+1}=3\times2^k+2(-1)^{k+1}}$

which is what we needed. QED

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3 years ago

PLEASE HELP What is the rule for the sequence with the first four terms below?

ivolga24 [154]

Answer:

A) [First try is correct for all U1 to U4]

Step-by-step explanation:

1. Trying each option

(let x = 1,2,3,4) since it goes from U1 to U4 (Geometric sequence)

A.) 10(-4/5)^x

10(-4/5)^1 = -8

10(-4/5)^2 = 6.4

10(-4/5)^3 = -5.12

10(-4/5)^4 = 4.096

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3 years ago