Stuck (2 Viewers)

Speed6 · Mar 29, 2015

Can someone please help me with the last part I am doing?

Ekman · Mar 29, 2015

Speed6 said:
Can someone please help me with the last part I am doing?

Just to continue your working out:

$10A -5\times3^k + 3\times3^k$

$10A - 2\times3^k$

$= 2(5A -3^k) \therefore $ For $ n=k+1 $ it is always divisible by 2$$

Speed6 · Mar 29, 2015

So how did you get the -2x3^k part man?

Kind of confused...

Ekman · Mar 29, 2015

Speed6 said:
So how did you get the -2x3^k part man?

Kind of confused...

$-5\times3^k + 3\times3^k = 3^k(-5+3) = 3^k\times -2$

Speed6 · Mar 29, 2015

Ekman said:
$-5\times3^k + 3\times3^k = 3^k(-5+3) = 3^k\times -2$

Thanks, got it now.

Speed6 · Mar 29, 2015

Ekman · Mar 29, 2015

Speed6 said:

To continue your working out:

$5\times5^k (4k+5) +3 = 5\times5^k(4k+1 + 4) +3$

$Assumption should have been established as: $ 5^k(4k+1) +3= 4m $ where m is an integer$

$\therefore 5\times5^k(4k+1 + 4) +3 = 5\times5^k(4k+1) + 3 + 20\times5^k = 4m + 20\times5^k$

$=4(m+5^k\times5) \therefore $ for $ n=k+1 $ it is always divisible by 4$$

Speed6 · Mar 29, 2015

Ekman, it's always the last part, could you help me out with this as well man?

(I might be doing the last part wrong, please check)

Speed6 · Mar 29, 2015

Oh I am doing it wrong, a very silly mistake lol. The indices and multiplication part is def wrong

Ekman · Mar 29, 2015

Speed6 said:
Ekman, it's always the last part, could you help me out with this as well man?

(I might be doing the last part wrong, please check)

For inequality questions, you cant substitute, so you must manipulate and show the inequality:

$$Assume :$ 7^k \geq 3^k + 4^k \therefore 7^k -3^k -4^k \geq 0$

$$ Test for $ n=k+1 : 7\times7^k \geq 3\times3^k + 4\times4^k$

$\therefore 7\times7^k - 3\times3^k - 4\times4^k \geq 0 \Rightarrow 7\times7^k - 7\times3^k + 4\times3^k - 7\times4^k +3\times4^k \geq 0$

$\therefore 7(7^k-3^k-4^k) + 4\times3^k + 3\times4^k \geq 0$

$$ Since: $ 7(7^k-3^k-4^k) \geq 0 $ (As assumed before) and $ 4\times3^k + 3\times4^k \geq 0$

$\therefore n=k+1 $ works for the inequality$$

Speed6 · Mar 29, 2015

Ok, that's something I need to add to my skillset.

Speed6 · Mar 29, 2015

Ekman · Mar 29, 2015

Speed6 said:

Im going to go ahead and skip the first step where you sub in n=1 or n=2.

$$ Assume for $ n=k: 1^2 + 3^2+5^2 +... + (2k-1)^2 = \frac{1}{3} k (2k-1)(2k+1)$

$$ Prove for $ n=k+1: 1^2 + 3^2 +... + (2k-1)^2 + (2(k+1)-1)^2 =\frac{1}{3} (k+1) (2k+1)(2k+3)$

$LHS=n=k+1: 1^2 + 3^2 +... + (2k-1)^2 + (2(k+1)-1)^2 = \frac{1}{3} k (2k-1)(2k+1) + (2k+1)^2 $ (From initial assumption)$$

$= (2k+1)(\frac{2k^2-k}{3} + 2k+1) = (2k+1)(\frac{2k^2+5k+3}{3})$

$$ Factorise: $ 2k^2 + 5k+3 = (k+1)(2k+3)$

$\therefore (2k+1)(\frac{2k^2+5k+3}{3})=\frac{1}{3} (k+1) (2k+1)(2k+3)$

$Since $ LHS = RHS $ $ n=k+1 $ works$

Speed6 · Mar 29, 2015

Ekman · Mar 29, 2015

Speed6 said:

Let BQR = a
a=RPQ (angle in alternate segment)
a=QRS (alternate angles in parallel lines)
therefore since QRS = QPS (angle in same segment)
QPS = a
Thus since QPS = RPQ, segment QP bisects SPR

Speed6 · Mar 29, 2015

Speed6 · Mar 29, 2015

InteGrand · Mar 29, 2015

Speed6 said:

angle DEC = angle DCE (base angles of isosceles triangle)

⇒ angle BEC = angle ECB + angle DCB
⇒ angle ECB = angle BEC – angle DCB
= angle BEC – angle EAC (alternate segment theorem)
= angle ECA, since angle ECA + angle EAC = angle BEC (external angle of triangle EAC).

Hence EC bisects angle BCA. Q.E.D.

Speed6 · Mar 29, 2015

Thanks Integrand

FlyingGullible · Mar 29, 2015

Let BCD = a
ECD = CED (base anges of isosceles triangle). Let this = b.
Therefore, since BCE = ECD - BCD
then BCE = b - a.

Also, by exterior angle triangle in triangle AEC:
ACE + EAC = CED
EAC = a (alternate segment theorem)
Therefore ACE = CED - EAC = b - a

So ACE = BCE. Therefore, it follows that EC bisects BCA.

Oh woops, looks like I got beaten to it.

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