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3 years ago

Evaluate the following expression using the values given: (1 point)

Mathematics

1 answer:

garik1379 [7]3 years ago

4 0

3(3)^2 - (-2)^3 - (-2)^3 + 5
27 - (-8) - (-8) +5
35 - (-8) + 5
43 + 5
48

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Prove by mathematical induction that 1+2+3+...+n= n(n+1)/2 please can someone help me with this ASAP. Thanks

Iteru [2.4K]

Let

$P(n):\ 1+2+\ldots+n = \dfrac{n(n+1)}{2}$

In order to prove this by induction, we first need to prove the base case, i.e. prove that P(1) is true:

$P(1):\ 1 = \dfrac{1\cdot 2}{2}=1$

So, the base case is ok. Now, we need to assume $P(n)$ and prove $P(n+1)$ .

$P(n+1)$ states that

$P(n+1):\ 1+2+\ldots+n+(n+1) = \dfrac{(n+1)(n+2)}{2}=\dfrac{n^2+3n+2}{2}$

Since we're assuming $P(n)$ , we can substitute the sum of the first n terms with their expression:

$\underbrace{1+2+\ldots+n}_{P(n)}+n+1 = \dfrac{n(n+1)}{2}+n+1=\dfrac{n(n+1)+2n+2}{2}=\dfrac{n^2+3n+2}{2}$

Which terminates the proof, since we showed that

$P(n+1):\ 1+2+\ldots+n+(n+1) =\dfrac{n^2+3n+2}{2}$

as required

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2 years ago

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I don't understand what you mean but this might be the answer
${6}^{100}$

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26.2% of smokers,when given this treatment, would refrain from smoking for at least 6 months.

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In this problem, we have that:

935 smokers received the treatment(nicotine patch).

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