The value of 1+\left(1+\frac{1}{3}\right)\frac{1}{4}+ \left(1+\frac{1}{3}+\frac{1}{9}\right)\frac{1}{16}+ \left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}\right)\frac{1}{64}$ + \ldots \ldots , is
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Official Correct Answer: C. The question is asking for the value of this series:
1 + (1 + 1/3)(1/4) + (1 + 1/3 + 1/9)(1/16) + (1 + 1/3 + 1/9 + 1/27)(1/64) + ... and so on.
The options are:
A) 27/12
B) 15/8
C) 16/11
D) 15/13
Hmm, okay. So it's an algebra problem involving series and progressions. Let me try to break it down.
First, Note that each term in the series after the first one has two parts: a sum of fractions and a fraction that's a power of 1/4. Let me write out the first few terms to see the pattern more clearly.
Term 1: 1
Term 2: (1 + 1/3) * (1/4)
Term 3: (1 + 1/3 + 1/9) * (1/16)
Term 4: (1 + 1/3 + 1/9 + 1/27) * (1/64)
And so on.
So, each term after the first is the sum of a geometric series multiplied by another term which is also a geometric series. Let me confirm that.
Looking at the sum part: 1 + 1/3 + 1/9 + 1/27 + ... This is a geometric series with first term 1 and common ratio 1/3. Similarly, the multiplier part: 1/4, 1/16, 1/64, ... This is also a geometric series with first term 1/4 and common ratio 1/4.
Wait, actually, the multiplier is 1/4, 1/16, 1/64, which is (1/4)^1, (1/4)^2, (1/4)^3, etc. So, the multiplier is (1/4)^n where n starts at 1.
Similarly, the sum inside each term is the sum of the first n terms of a geometric series with a=1 and r=1/3. The sum of the first n terms of a geometric series is given by S_n = (1 - r^n)/(1 - r). So, for each term, the sum part is S_n = (1 - (1/3)^n)/(1 - 1/3) = (1 - (1/3)^n)/(2/3) = (3/2)(1 - (1/3)^n).
Therefore, each term after the first can be written as:
Term n: [ (3/2)(1 - (1/3)^n) ] * (1/4)^n
But wait, the first term is 1, which is when n=0? Or maybe n starts at 1. Let me check.
Term 1: n=1: [ (3/2)(1 - 1/3) ] * (1/4)^1 = (3/2)(2/3) * (1/4) = (1) * (1/4) = 1/4. But in the series, the second term is 1/4, so maybe n starts at 1 for the terms after the first.
Wait, the series is 1 + term2 + term3 + term4 + ..., so the first term is 1, and then starting from term2, n=1, term3 n=2, etc.
So, the entire series can be written as:
Sum = 1 + Σ [ (3/2)(1 - (1/3)^n) * (1/4)^n ] from n=1 to infinity.
So, let me write that as:
Sum = 1 + (3/2) Σ [ (1 - (1/3)^n ) * (1/4)^n ] from n=1 to ∞.
I can split this sum into two separate sums:
Sum = 1 + (3/2) [ Σ (1/4)^n - Σ (1/3)^n * (1/4)^n ] from n=1 to ∞.
Compute each of these sums separately.
First, compute Σ (1/4)^n from n=1 to ∞. That's a geometric series with a = 1/4 and r = 1/4. The sum is a / (1 - r) = (1/4) / (1 - 1/4) = (1/4) / (3/4) = 1/3.
Next, compute Σ (1/3)^n * (1/4)^n from n=1 to ∞. That's the same as Σ (1/12)^n from n=1 to ∞, since (1/3)*(1/4) = 1/12. So, this is another geometric series with a = 1/12 and r = 1/12. The sum is (1/12) / (1 - 1/12) = (1/12) / (11/12) = 1/11.
So, putting it back together:
Sum = 1 + (3/2) [ (1/3) - (1/11) ].
Compute the difference inside the brackets:
(1/3) - (1/11) = (11 - 3)/33 = 8/33.
So, Sum = 1 + (3/2)*(8/33).
Simplify (3/2)*(8/33):
3 and 33 have a common factor of 3: 3/33 = 1/11.
So, (3/2)*(8/33) = (1/2)*(8/11) = 4/11.
Therefore, Sum = 1 + 4/11 = 15/11.
Wait, but looking at the options, 15/11 isn't one of them. The options are 27/12, 15/8, 16/11, 15/13.
Hmm, did I make a mistake somewhere?
First, I wrote the sum as 1 + Σ [ (3/2)(1 - (1/3)^n) * (1/4)^n ] from n=1 to ∞.
Then, I split it into two sums: (3/2)[Σ (1/4)^n - Σ (1/12)^n ].
Computed Σ (1/4)^