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3 years ago

Log base 2 (1/4) ...?

1 answer:

5 0

Log ( base 2 ) ( 1 / 4 ) =
= log ( base 2 ) ( 2 ^(-2 ) ) =
= - 2 log ( base 2 ) 2 = ( because : log x^n = n log x )
= ( - 2 ) * 1 = ( because: log (base x) x = 1 )
= - 2

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Which symbol makes the comparisons true ? Complete the comparison -6 ? 3

Usimov [2.4K]

<h3>Hello!</h3>

$======================================$

Negative numbers are always less than positive ones.

Do you remember the symbol for "less than"? That's right, "<" :)

Hence,

-6<3

Because

-6 is negative (the - sign tells us that)

3 is positive

$===================================$

<h3>Notes:</h3>

Hope everything is clear.
Let me know if you have any questions!

<h3>Answered by:</h3>

~DiamondS~

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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.

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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 <em>S</em>₁ = 1.

• Assume the closed form is correct for all <em>n</em> up to <em>n</em> = <em>k</em>. 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 <em>n</em> = <em>k</em> + 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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