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3 years ago

Write down an expression for the nth term of the following sequence 5, 11, 17, 23, ...

Mathematics

2 answers:

AysviL [449]3 years ago

8 0

Answer:

$a_{n}$ = 6n - 1

Step-by-step explanation:

There is a common difference d between consecutive terms in the sequence

d = 11 - 5 = 17 - 11 = 23 - 17 = 6

This indicates the sequence is arithmetic with n th term

$a_{n}$ = a₁ + (n - 1)d

where a₁ is the first term and d the common difference

Here a₁ = 5 and d = 6 , then

$a_{n}$ = 5 + 6(n - 1) = 5 + 6n - 6 = 6n - 1

Scorpion4ik [409]3 years ago

4 0

6n-1
Because:
6n= 6,12,18,24
6n-1=6x1-1=5,6x2-1=11,6x3-1=17 and 6x4-1=23

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Find an explicit formula for the following sequence, an which starts with a1=2/3. 2/3,3/4,4/5,5/6,…

Flura [38]

Answer:

Step-by-step explanation:

a1=2/3

sequence is 2/3,3/4,4/5,...

for numerator a1=2

d=3-2=1

numerator of nth term=a1+(n-1)d=2+(n-1)×1=2+n-1=n+1

denominator = 1 more than numerator=n+1+1=n+2

so an=(n+1)/(n+2)

or for denominator a1=3,d=4-3=1

denominator of nth term=3+(n-1)×1=3+n-1=n+2

an=(n+1)/(n+2)

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3 years ago

y′′ −y = 0, x0 = 0 Seek power series solutions of the given differential equation about the given point x 0; find the recurrence

sukhopar [10]

Let

$\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = a_0 + a_1x + a_2x^2 + \cdots$

Differentiating twice gives

$\displaystyle y'(x) = \sum_{n=1}^\infty na_nx^{n-1} = \sum_{n=0}^\infty (n+1) a_{n+1} x^n = a_1 + 2a_2x + 3a_3x^2 + \cdots$

$\displaystyle y''(x) = \sum_{n=2}^\infty n (n-1) a_nx^{n-2} = \sum_{n=0}^\infty (n+2) (n+1) a_{n+2} x^n$

When x = 0, we observe that y(0) = a₀ and y'(0) = a₁ can act as initial conditions.

Substitute these into the given differential equation:

$\displaystyle \sum_{n=0}^\infty (n+2)(n+1) a_{n+2} x^n - \sum_{n=0}^\infty a_nx^n = 0$

$\displaystyle \sum_{n=0}^\infty \bigg((n+2)(n+1) a_{n+2} - a_n\bigg) x^n = 0$

Then the coefficients in the power series solution are governed by the recurrence relation,

$\begin{cases}a_0 = y(0) \\ a_1 = y'(0) \\\\ a_{n+2} = \dfrac{a_n}{(n+2)(n+1)} & \text{for }n\ge0\end{cases}$

Since the n-th coefficient depends on the (n - 2)-th coefficient, we split n into two cases.

• If n is even, then n = 2k for some integer k ≥ 0. Then

$k=0 \implies n=0 \implies a_0 = a_0$

$k=1 \implies n=2 \implies a_2 = \dfrac{a_0}{2\cdot1}$

$k=2 \implies n=4 \implies a_4 = \dfrac{a_2}{4\cdot3} = \dfrac{a_0}{4\cdot3\cdot2\cdot1}$

$k=3 \implies n=6 \implies a_6 = \dfrac{a_4}{6\cdot5} = \dfrac{a_0}{6\cdot5\cdot4\cdot3\cdot2\cdot1}$

It should be easy enough to see that

$a_{n=2k} = \dfrac{a_0}{(2k)!}$

• If n is odd, then n = 2k + 1 for some k ≥ 0. Then

$k = 0 \implies n=1 \implies a_1 = a_1$

$k = 1 \implies n=3 \implies a_3 = \dfrac{a_1}{3\cdot2}$

$k = 2 \implies n=5 \implies a_5 = \dfrac{a_3}{5\cdot4} = \dfrac{a_1}{5\cdot4\cdot3\cdot2}$

$k=3 \implies n=7 \implies a_7=\dfrac{a_5}{7\cdot6} = \dfrac{a_1}{7\cdot6\cdot5\cdot4\cdot3\cdot2}$

so that

$a_{n=2k+1} = \dfrac{a_1}{(2k+1)!}$

So, the overall series solution is

$\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = \sum_{k=0}^\infty \left(a_{2k}x^{2k} + a_{2k+1}x^{2k+1}\right)$

$\boxed{\displaystyle y(x) = a_0 \sum_{k=0}^\infty \frac{x^{2k}}{(2k)!} + a_1 \sum_{k=0}^\infty \frac{x^{2k+1}}{(2k+1)!}}$

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3 years ago

3. Determine whether the set of numbers can be the measures of the sides of a triangle. If so, classify the triangle as acute, o

julia-pushkina [17]

Answer:

all of them are acute

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3 years ago

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How do you divide 270 into 3000

chubhunter [2.5K]

3000
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3 years ago

6. Gabriella invests $6,500 into a savings account that earns 4.5% interest, and she is investing the money for 5 years without

sdas [7]

Answer:

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Step-by-step explanation:

6500

x.045

=292.50 per year

x. 5 years

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3 years ago