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# If (a+b)^x=a^x + y(a^(x-1)*b^(x-4)) + z(a^(x-2)*b^(x-3)) + z(a^(x-3)*^

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Joined: 02 Sep 2009
Posts: 58366
If (a+b)^x=a^x + y(a^(x-1)*b^(x-4)) + z(a^(x-2)*b^(x-3)) + z(a^(x-3)*^  [#permalink]

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06 Nov 2014, 09:09
4
28
00:00

Difficulty:

95% (hard)

Question Stats:

46% (02:17) correct 54% (02:25) wrong based on 224 sessions

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Tough and Tricky questions: Exponents.

If $$(a+b)^x=a^x + y(a^{(x-1)}*b^{(x-4)}) + z(a^{(x-2)}*b^{(x-3)}) + z(a^{(x-3)}*b^{(x-2)}) + y(a^{(x-4)}*b^{(x-1)}) + b^x$$, what is the value of yz?

A. 24
B. 30
C. 36
D. 42
E. 50

Kudos for a correct solution.

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Re: If (a+b)^x=a^x + y(a^(x-1)*b^(x-4)) + z(a^(x-2)*b^(x-3)) + z(a^(x-3)*^  [#permalink]

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09 Nov 2014, 20:20
10
3
Standard formula is:

$$(a+b)^n = n_{C_0} a^n b^0 + n_{C_1} a^{n-1} b^1 + n_{C_2} a^{n-2} b^2 +........... + n_{C_n} a^0 b^n$$

By comparing the standard formula with the given equation, its confirmed that x = n = 5

$$y = 5_{C_1} = \frac{5!}{4!1!} = 5$$

$$z = 5_{C_2} = \frac{5!}{2!3!} = 10$$

yz = 5*10 = 50

There is also an exponential pyramid. Please refer below:
Attachment:

pyra.png [ 11.15 KiB | Viewed 3900 times ]

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##### General Discussion
Senior Manager
Joined: 13 Jun 2013
Posts: 266
Re: If (a+b)^x=a^x + y(a^(x-1)*b^(x-4)) + z(a^(x-2)*b^(x-3)) + z(a^(x-3)*^  [#permalink]

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07 Nov 2014, 05:44
2
1
Bunuel wrote:

Tough and Tricky questions: Exponents.

If $$(a+b)^x=a^x + y(a^{(x-1)}*b^{(x-4)}) + z(a^{(x-2)}*b^{(x-3)}) + z(a^{(x-3)}*b^{(x-2)}) + y(a^{(x-4)}*b^{(x-1)}) + b^x$$, what is the value of yz?

A. 24
B. 30
C. 36
D. 42
E. 50

Kudos for a correct solution.

before solving this question. let's look into the very simple formula of (a+b)^2
(a+b)^2=a^2+b^2+2ab

if we see, the maximum power (or degree) of each term in the expression is 2

so, by using same corollary we can assume that, the highest power of term in the expression (a+b)^x will be x

thus, by considering the second term we have (x-1)+(x-4)=x
or x=5
thus the value of x=5. now the expression given in the question becomes.
$$(a+b)^5=a^5 + y(a^{(4)}*b^{(1)}) + z(a^{(3)}*b^{(2)}) + z(a^{(2)}*b^{(3)}) + y(a^{(1)}*b^{(4)}) + b^5$$

also, (a+b)^3= (a+b)(a+b)^2
= (a+b)(a^2+b^2+2ab)
=a^3+ab^2+2(a^2)b+(a^2)b+b^3+2ab^2
=a^3+b^3+3(a^2)b+3ab^2

(a+b)^5= (a+b)^3(a+b)^2
= (a^3+b^3+3(a^2)b+3ab^2)(a^2+b^2+2ab)----------------------1)
also, if we look at the given expression in the question. we will notice that we only have to focus on the co-eff. of terms containing a^4b and a^3b^2

so, let's just focus only on these terms in 1)
=3(a^4)b +2a^4b +a^3b^2+6a^3b^2+3a^3b^2
=5a^4b+10a^3b^2

thus y=5 and z=10
and yz=5(10)=50

p.s. this question can also be solved by using binomial theorem.
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Re: If (a+b)^x=a^x + y(a^(x-1)*b^(x-4)) + z(a^(x-2)*b^(x-3)) + z(a^(x-3)*^  [#permalink]

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07 Nov 2014, 06:35
1
1
Bunuel wrote:

Tough and Tricky questions: Exponents.

If $$(a+b)^x=a^x + y(a^{(x-1)}*b^{(x-4)}) + z(a^{(x-2)}*b^{(x-3)}) + z(a^{(x-3)}*b^{(x-2)}) + y(a^{(x-4)}*b^{(x-1)}) + b^x$$, what is the value of yz?

A. 24
B. 30
C. 36
D. 42
E. 50

Kudos for a correct solution.

If we substitute x =2, we get
$$(a+b)^2 = a^2 + 2ab + b^2$$

If we substitute x = 3, we get
$$(a+b)^3 = (a+b)(a+b)^2 = a^3 + 3b(a)^2 + 3a(b)^2 + b^3$$

Now if substitute x=5, we get
$$(a+b)^5 = (a+b)^2 * (a+b)^3$$

$$(a+b)^5 = a^5 + [5*b*(a)^4] + [10(a)^3*(b)^2] + [10*(a)^2* (b)^3] + [5*a*(b)^4] + b^5$$

$$(a+b)^5 = a^5 + 5[(a)^4b] + 10[(a)^3(b)^2] + 10[(a)^2 (b)^3] + 5[a(b)^4] + b^5$$

From above equation, we get
x=5 ; (x-1) = 4 ; (x-2) = 3 ; (x-3) = 2
and
y=5 ; z = 10

therefore, yz = 50

I would like to know if there is a faster way to solve than doing the algebra part.
Manager
Status: I am not a product of my circumstances. I am a product of my decisions
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Location: India
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Re: If (a+b)^x=a^x + y(a^(x-1)*b^(x-4)) + z(a^(x-2)*b^(x-3)) + z(a^(x-3)*^  [#permalink]

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09 Nov 2014, 20:51
PareshGmat wrote:
Standard formula is:

$$(a+b)^n = n_{C_0} a^n b^0 + n_{C_1} a^{n-1} b^1 + n_{C_2} a^{n-2} b^2 +........... + n_{C_n} a^0 b^n$$

By comparing the standard formula with the given equation, its confirmed that x = n = 5

$$y = 5_{C_1} = \frac{5!}{4!1!} = 5$$

$$z = 5_{C_2} = \frac{5!}{2!3!} = 10$$

yz = 5*10 = 50

There is also an exponential pyramid. Please refer below:
Attachment:
pyra.png

Thats a good solution Paresh but can you tell me that where can i find a write up about this Formula.
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Joined: 27 Dec 2012
Posts: 1749
Location: India
Concentration: General Management, Technology
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Re: If (a+b)^x=a^x + y(a^(x-1)*b^(x-4)) + z(a^(x-2)*b^(x-3)) + z(a^(x-3)*^  [#permalink]

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10 Nov 2014, 02:39
Ashishmathew01081987 wrote:
PareshGmat wrote:
Standard formula is:

$$(a+b)^n = n_{C_0} a^n b^0 + n_{C_1} a^{n-1} b^1 + n_{C_2} a^{n-2} b^2 +........... + n_{C_n} a^0 b^n$$

By comparing the standard formula with the given equation, its confirmed that x = n = 5

$$y = 5_{C_1} = \frac{5!}{4!1!} = 5$$

$$z = 5_{C_2} = \frac{5!}{2!3!} = 10$$

yz = 5*10 = 50

There is also an exponential pyramid. Please refer below:
Attachment:
pyra.png

Thats a good solution Paresh but can you tell me that where can i find a write up about this Formula.

Thanks sir. If you are asking about "how the formula is derived", then I'm not sure.....

I came across this in CBSE Class XII book way back
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Joined: 28 Jan 2017
Posts: 49
Re: If (a+b)^x=a^x + y(a^(x-1)*b^(x-4)) + z(a^(x-2)*b^(x-3)) + z(a^(x-3)*^  [#permalink]

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24 May 2017, 14:50
Bunuel wrote:

Tough and Tricky questions: Exponents.

If $$(a+b)^x=a^x + y(a^{(x-1)}*b^{(x-4)}) + z(a^{(x-2)}*b^{(x-3)}) + z(a^{(x-3)}*b^{(x-2)}) + y(a^{(x-4)}*b^{(x-1)}) + b^x$$, what is the value of yz?

A. 24
B. 30
C. 36
D. 42
E. 50

Kudos for a correct solution.

Hi Bunuel,

Does GMAT ask questions of this difficulty.
This can be solved only by the people who are in constant touch with Math. Moreover, 2 mins do not seems to be enough for this question.
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Joined: 23 Sep 2014
Posts: 6
Re: If (a+b)^x=a^x + y(a^(x-1)*b^(x-4)) + z(a^(x-2)*b^(x-3)) + z(a^(x-3)*^  [#permalink]

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18 Jun 2017, 22:51
PareshGmat wrote:
Ashishmathew01081987 wrote:
PareshGmat wrote:
Standard formula is:

$$(a+b)^n = n_{C_0} a^n b^0 + n_{C_1} a^{n-1} b^1 + n_{C_2} a^{n-2} b^2 +........... + n_{C_n} a^0 b^n$$

By comparing the standard formula with the given equation, its confirmed that x = n = 5

$$y = 5_{C_1} = \frac{5!}{4!1!} = 5$$

$$z = 5_{C_2} = \frac{5!}{2!3!} = 10$$

yz = 5*10 = 50

There is also an exponential pyramid. Please refer below:
Attachment:
pyra.png

Thats a good solution Paresh but can you tell me that where can i find a write up about this Formula.

Thanks sir. If you are asking about "how the formula is derived", then I'm not sure.....

I came across this in CBSE Class XII book way back

Google binomial theorem and its proof.. You will get it easily
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Posts: 13162
Re: If (a+b)^x=a^x + y(a^(x-1)*b^(x-4)) + z(a^(x-2)*b^(x-3)) + z(a^(x-3)*^  [#permalink]

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17 Jan 2019, 19:24
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Re: If (a+b)^x=a^x + y(a^(x-1)*b^(x-4)) + z(a^(x-2)*b^(x-3)) + z(a^(x-3)*^   [#permalink] 17 Jan 2019, 19:24
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