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2 years ago

F(n) = 4(n-1) + 7fine the first three(3) terms in the sequence show all work

Mathematics

2 answers:

vladimir1956 [14]2 years ago

8 0

........................................

lubasha [3.4K]2 years ago

4 0

Answer:

f(n) = 4(n-1) + 7

First term

f(1) = 4(1-1) +7 = 0 + 7 = 7

Second term

f(2) = 4(2-1) + 7

= 4(1) + 7 = 4 + 7 =11

Third term

f(3) = 4(3-1) + 7

= 4(2) + 7

= 8 + 7 = 15

Therefore the first three terms of the sequence are 7, 11 and 15.

Hope this helps

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2 years ago

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$\lim_{n \to \infty} U_n =0$

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Step-by-step explanation:

Given series is an alternating series

∑ $(-1)^{n} \frac{n^{2} }{n^{3}+3 }$

Let $U_{n} = (-1)^{n} \frac{n^{2} }{n^{3}+3 }$

By using Leibnitz's rule

$U_{n} - U_{n-1} = \frac{n^{2} }{n^{3} +3} - \frac{(n-1)^{2} }{(n-1)^{3}+3 }$

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Uₙ-Uₙ₋₁ <0

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2 years ago

By taking  = 3.141 and

Whitepunk [10]

The value of the expression 1/√2 + π where the given parameters are π = 3.141 and √2 = 1.414 is 3.848

<h3>How to evaluate the expression?</h3>

The given parameters are

π = 3.141

√2 = 1.414

The expression to evaluate is given as:

1/√2 + π

Start by substituting π = 3.141 and √2 = 1.414 in the expressions 1/√2 + π

1/√2 + π = 1/1.414 + 3.141

Evaluate the quotient

1/√2 + π = 0.707 + 3.141

Evaluate the sum

1/√2 + π = 3.848

Hence, the value of the expression 1/√2 + π where the given parameters are π = 3.141 and √2 = 1.414 is 3.848

F(n) = 4(n-1) + 7fine the first three(3) terms in the sequence show all work​

F(n) = 4(n-1) + 7fine the first three(3) terms in the sequence show all work