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3 years ago

11

Long rationalise surds questions pls help

1 answer:

Katena32 [7]3 years ago

3 0

Answer:

Noah mistakenly added 25 and 7 in denominator instead of subtracting in the third step.

Step-by-step explanation:

Noah had to rationalize the denominator of $\frac{32}{5-\sqrt{7}}$

According to him,

$\frac{32}{5-\sqrt{7} = \frac{32(5+\sqrt{7}{(5-\sqrt{7})(5+\sqrt{7}$

Solving the nominator and denominator

$= \frac{160-32\sqrt{7}}{25+5\sqrt{7}-5\sqrt{7}-7}$

Noah did correctly till this point.

In the next step he add 25 and 7 in the denominator, whereas, it should be subtracted.

Hence,

Noah mistakenly added 25 and 7 in denominator instead of subtracting in the third step.

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Determine formula of the nth term 2, 6, 12 20 30,42

nalin [4]

Check the forward differences of the sequence.

If $\{a_n\} = \{2,6,12,20,30,42,\ldots\}$ , then let $\{b_n\}$ be the sequence of first-order differences of $\{a_n\}$ . That is, for n ≥ 1,

$b_n = a_{n+1} - a_n$

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and we can solve for $b_n$ in terms of $b_1=4$ :

$b_{n+1} = b_n + 2$

$b_{n+1} = (b_{n-1}+2) + 2 = b_{n-1} + 2\times2$

$b_{n+1} = (b_{n-2}+2) + 2\times2 = b_{n-2} + 3\times2$

and so on down to

$b_{n+1} = b_1 + 2n \implies b_{n+1} = 2n + 4 \implies b_n = 2(n-1)+4 = 2(n + 1)$

We solve for $a_n$ in the same way.

$2(n+1) = a_{n+1} - a_n \implies a_{n+1} = a_n + 2(n + 1)$

Then

$a_{n+1} = (a_{n-1} + 2n) + 2(n+1) \\ ~~~~~~~= a_{n-1} + 2 ((n+1) + n)$

$a_{n+1} = (a_{n-2} + 2(n-1)) + 2((n+1)+n) \\ ~~~~~~~ = a_{n-2} + 2 ((n+1) + n + (n-1))$

$a_{n+1} = (a_{n-3} + 2(n-2)) + 2((n+1)+n+(n-1)) \\ ~~~~~~~= a_{n-3} + 2 ((n+1) + n + (n-1) + (n-2))$

and so on down to

$a_{n+1} = a_1 + 2 \displaystyle \sum_{k=2}^{n+1} k = 2 + 2 \times \frac{n(n+3)}2$

$\implies a_{n+1} = n^2 + 3n + 2 \implies \boxed{a_n = n^2 + n}$

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2 years ago

Please help if you would like brianleist!!!

BartSMP [9]

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