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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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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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New post 06 Nov 2014, 09:09
4
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A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

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

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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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New post 09 Nov 2014, 20:20
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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

Answer = E

There is also an exponential pyramid. Please refer below:
Attachment:
pyra.png
pyra.png [ 11.15 KiB | Viewed 3900 times ]

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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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New post 07 Nov 2014, 05:44
2
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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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New post 07 Nov 2014, 06:35
1
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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

Answer is E.

I would like to know if there is a faster way to solve than doing the algebra part.
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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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New post 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

Answer = E

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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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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New post 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

Answer = E

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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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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New post 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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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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New post 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

Answer = E

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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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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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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