In the sequence an = (x + (1)n 1)(an1) and a1 = 1, which of the

Question

In the sequence  a_n = (x + (–1)^{(n –1)})(a_{n–1})  and  a_1 = 1 , which of the following expresses a4?

A.  x^2 – 1 
B.  x^3 – x^2 – x + 1 
C.  x^3 + x^2 + x – 1 
D.  x^4 – 2x^2 + 1 
E.  x^4 + 2x^2 – 1

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ravi1522 · Answer

Hello,
As a1=1
a2=x-1
a3=(x+1)*(X-1)
a4=(x^2-1)(X-1)
ie x^3–x^2–x+1

Option B is correct

AyushiTri · Answer

an=(x+(–1)^(n–1))(an–1) and a1=1 is given.

Therefore a2= (x+(–1)^(n–1)) a1 = (x-1) *1 = x-1

and so a3 = (x+(–1)^(n–1)) a2 = (x+(–1)^(3–1)) (x-1) = (x+1) (x-1) = x^2 - 1

Now a4 = (x+(–1)^(n–1)) a3 = (x-1) (x^2 - 1) = X^3 - x^2 - X + 1 Which is B.

Answer is B.

rohitz98 · Answer

Given  an=(x+(–1)^(n–1))(an–1) and a1=1

Hence a2 = (x - 1)(1) 
  a3 = (x+1)(x-1) 
  a4 = (x-1)(x+1)(x-1) 
 Simplifying we get  x^3 - x^2 - x + 1 Option B

ManifestDreamMBA · Answer

a_n = (x + (–1)^{(n –1)})(a_{n–1}) 
 a4 = (x + (–1)^{(4 –1)})a_3 
 a4 = (x –1)*a_3 
 a4 = (x –1)(x + (–1)^{(3 –1)})(a_{3–1}) 
 a4 = (x –1)(x + 1)*a_2 
 a4 = (x –1)(x + 1)(x + (–1)^{(2 –1)})(a_{2–1}) 
 a4 = (x –1)(x + 1)(x –1)*a_1 
 a4 = (x –1)(x + 1)(x –1)*1 
 a4 = (x^2 –1)(x –1) 
 a4 = x^3 – x^2 – x + 1

Dereno · Answer

Given a1 = 1

When n =2, we get a2 = (x-1)

When n = 3 , we get a3 = (x+1)*(x-1) = (x^2 - 1)

When n =4, we get a4 = (x-1)*(x+1)*(x-1) = (x-1)^2 * (x+1)

= (x^2+ 1 -2x) *(x+1)

= x^3 + x - 2*x^2 + x^2 + 1 - 2x

= x^3 - x^2 - x +1

B.  x^3 – x^2 – x + 1

In the sequence an = (x + (1)n 1)(an1) and a1 = 1, which of the

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GMAT Problem Solving (PS) Questions

In the sequence an = (x + (1)n 1)(an1) and a1 = 1, which of the

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