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15 Sep 2024, 02:09
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Bunuel
k/h = x^5
here we have to consider that the denominator h is
x+x^2+x^3+x^4+x^5
to obtain x^5 as the ratio we need
k = x^5 (x+x^2+x^3+x^4+x^5)
h = x+x^2+x^3+x^4+x^5
k = x^6 + x^7+x^8+x^9+x^10
if we assume k to be a value of the series An
than to obtain k from An, n must be equal to 7
A7 = x^6 + x^7+x^8+x^9+x^10
k/h = x^5
here we have to consider that the denominator h is
x+x^2+x^3+x^4+x^5
to obtain x^5 as the ratio we need
k = x^5 (x+x^2+x^3+x^4+x^5)
h = x+x^2+x^3+x^4+x^5
k = x^6 + x^7+x^8+x^9+x^10
if we assume k to be a value of the series An
than to obtain k from An, n must be equal to 7
A7 = x^6 + x^7+x^8+x^9+x^10
06 Oct 2024, 07:05
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Does this approach make sense?
An/x(1 + x(1 + x(1 + x(1 + x)))) = x^5
multiplying denominator to right side, we get:
An = x^5 * x(1 + x(1 + x(1 + x(1 + x))))
expand An using the definition given:
x^n – 1 + x^n + x^n + 1 + x^n + 2 + x^n + 3 = same RHS as above
degree(highest power) of LHS = x^n + 3
degree(highest power) of RHS = x^5 * x^5
n+3 = 5+5
n= 7
Note: ^ means raised to the power
An/x(1 + x(1 + x(1 + x(1 + x)))) = x^5
multiplying denominator to right side, we get:
An = x^5 * x(1 + x(1 + x(1 + x(1 + x))))
expand An using the definition given:
x^n – 1 + x^n + x^n + 1 + x^n + 2 + x^n + 3 = same RHS as above
degree(highest power) of LHS = x^n + 3
degree(highest power) of RHS = x^5 * x^5
n+3 = 5+5
n= 7
Note: ^ means raised to the power
adityaprateek15
Joined: 26 May 2023
Last visit: 13 May 2026
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23 Apr 2025, 07:31
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If you expand: x(1+x(1+x(1+x(1+x)))), you'll get: x+x^2+x^3+x^4+x^5
What you've been asked: For what value of An:
An/x+x^2+x^3+x^4+x^5 = x^5
An = X^5( x+x^2+x^3+x^4+x^5)
An = x^6+x^7+x^8+x^9+x^10 - I
What we have: An = X^(n-1) + X^n + X^(n+1) + X^(n+2) + X^(n+3) - II
If you compare I and II,
n-1=6 => n=7
What you've been asked: For what value of An:
An/x+x^2+x^3+x^4+x^5 = x^5
An = X^5( x+x^2+x^3+x^4+x^5)
An = x^6+x^7+x^8+x^9+x^10 - I
What we have: An = X^(n-1) + X^n + X^(n+1) + X^(n+2) + X^(n+3) - II
If you compare I and II,
n-1=6 => n=7
14 Jul 2025, 03:07
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I followed this logic :
- the question mentions we get x^5 when dividing the A by the denominator
- the max power of the denominator is x^5
- meaning that the numerator must have a term with x^10
When n=7, A = (x^6)+(x^7) + (x^8)+ (x^9) + (x^10)
Hence, ans B.
Pls let me know if this method could go wrong
- the question mentions we get x^5 when dividing the A by the denominator
- the max power of the denominator is x^5
- meaning that the numerator must have a term with x^10
When n=7, A = (x^6)+(x^7) + (x^8)+ (x^9) + (x^10)
Hence, ans B.
Pls let me know if this method could go wrong
