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

Question

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

PareshGmat · Accepted Answer

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: pyra.png

manpreetsingh86 · Answer

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.

Ashishmathew01081987 · Answer

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 +   +   +   +   + b^5

(a+b)^5 = a^5 + 5  + 10  + 10  + 5  + 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.

Ashishmathew01081987 · Answer

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

PareshGmat · Answer

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

CantDropThisTime · Answer

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.

laasshetty · Answer

Google binomial theorem and its proof.. You will get it easily

Crytiocanalyst · Answer

The variable expansion of the terms is the binominal expansion in 5

y , z are 5c1 and 5c2 giving us vale of 5 and 10

THerefore y*z = 50 IMO E

gowthamkanirajan · Answer

thanks for the simplest solution for complicated problem. May i know how you figured that out ?

Posted from my mobile device

bumpbot · Answer

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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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