Which of the following is closest to

Difference of squares:

(1.03^4-1) = (1.03^2+1)*(1.03^2-1) =

(1.03^2+1)(1.03+1)(1.03-1)

100(.03)(2.03)(1.03^2+1)

Now, 1.03^2 is just a small amount more than 1.03 and the answers are far apart enough to estimate as:

(3)(2)(2) = 12

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­Which of the following is closest to \(100((1 + 0.03)^4 - 1)\)?

When we multiply a number by \(1.0x\), the product is \(x\) percent greater than the original number.

So, each time we multiply \(1.03\) by itself, the result is \(3\) percent greater.

So, \(1.03 × 1.03\) is three percent greater than \(1.03\), or about \(1.06\).

\(1.03^3\) would then be about \(1.09\).

\(1.03^4\) is about \(1.12\).

So, \((1 + 0.03)^4 - 1 ≈ 0.12\)

\(0.12 × 100 = 12\)

A. 0.001

B. 0.126

C. 12.600

D. 62.012

E. 112.60

So, the correct answer is C.­

Different Approach using Binomial Theorem, which is also useful to estimate compound growth in consulting case interview for example:
In case you only know the binomial theorem for (x+y)^2, in this case you can do the following (this is basically a prove for the approximation of @gmatophobia):

Initial Formula: 100((1+0.03)^4-1))
1. Step Rewrite to ^2: 100((1+0.03)^2*(1+0.03)^2-1))
2. Apply binomial theorem: (1+0.03)^2 = 1^2 + 2*0.03 + 0.03^2 = 1 + 0.06 + 0.0009; 0.0009 is so small that you can drop it and there for it is approximately 1.06.
3. So it is 100(1.06*1.06-1). Either you can use long multiplication if you do not recognize difference in squares as you could have also done initially (as mentioned by @Regor60) or apply the same process again: (1+0.06)^2 = 1^2 + 2*0.06 + 0.06^2 = 1 + 0.12 + 0.0036 = 1.1236
4. 100(1.1236-1) = 100(0.1236) = 12.36; Therefore C

The binomial theorem does not only apply to (x+y)^2 but to any (x+y)^n. Here you can see the generalized version of the binomial theorem on Wikipedia.
Based on the solution above, we can infer that the terms further to the right become smaller and not too relevant. For any (x+y)^n the first three terms have the most impact in the binomial theorem. Therefore you can approximate (x+y)^n, based on the binomial theorem, by: x^n + n*y + n*(n-1)/2 * y^2

For this problem it would be: (1+0.03)^4 ≈ 1^4 + 4*0.03 + 4*(4-1)/2 * 0.03^2 = 1 + 0.12 + 6 * 0.0009 = 1.1254
100(1.1254-1) = 12.54

Applied to a consulting case or estimating compound growth in general also on the GMAT:
What will the 100 Million Bubble Gum Market currently (in 2024) be in 2029 if it grew at 5% per year?
1. Translate general compound interest/growth formula: 100 Million * (1+0.05)^5
2. Estimate compounding factor (1+0.05)^5 using the binomial theorem approach above: (1+0.05)^5 ≈ 1^5 + 5 * 0.05 + 5*(5-1)/2 * 0.05^2 = 1 + 0.25 + 10 * 0.0025 = 1.275
3. 100 Million * 1.275 = 127.5 Million. The accurate number is 127.628 Million, which is pretty close for such an easy calculation.

Let's do another one using compound interest:
100$ is compounded at 20% for 5 year
1. Translate to compound growth formula: 100*(1+0.2)^5
2. Estimate compounding factor using binomial theorem approximation: 1^5 + 5 * 0.2 + 5*(5-1)/2 * 0.2^2 = 1 + 1 + 10 * 0.04 = 2.4
3. 100*2.4 = 240; The actual number is 248.832, close enough for any GMAT question or case interview

Shortcut: Exponents of decimal < 1 such as 0.03^2
Also easy way to take the exponent of a decimal (such as 0.03^2 or 0.03^4) is to apply the decimal to the non-zero digits, in this case 3, so 3^2 = 9, multiply the exponent by the number of digits after the dot (0.03 has 2 digits after the dot), so 2*2 = 4, which are the total decimal digits. Combine those two then to 0.0009 (count 4 digits after the dot of which 1 is the square of 3 (9)). In a nutshell, (1) take the exponent of the non-zero digit, (2) multiply the exponent by the number of decimal digits, (3) subtract the number of digits of the results of (1) from the number of digits of the results from (2) to obtain the number of leading zeros of the new decimal in front of the results of (1)

Thank you gmatophobia great way indeed..kudos !! you are a genius !!

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