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The key to this problem is realizing that 3 + 3 + 3 = 9 = 3^2. Let us rewrite the prompt accordingly: 3^2 + 2 × 3^2 + 2 × 3^3 + 2 × 3^4 + 2 × 3^5 + 2 × 3^6 + 2 × 3^7.

Now, since this equation contains exponents, we should try to manipulate the terms so that we can use the exponent rules we know. We can rewrite the first three terms as 3^2 + 3^2 + 3^2. This is equivalent to 3(3^2) = 3^1(3^2) = 3^3.

The equation can thus be simplified: 3^3 + 2 × 3^3 + 2 × 3^4 + 2 × 3^5 + 2 × 3^6 + 2 × 3^7.

Since 3 × 3^3 = 3^4, we can simplify further: 3^4 + 2 × 3^4 + 2 × 3^5 + 2 × 3^6 + 2 × 3^7.

We repeat this procedure until we get: 3 × 3^7 = 38.

We have the sum of 9 terms. Now, if all terms were equal to the largest term 2*3^7 we would have: sum=9*(2*3^7)=2*3^9=~3^10, so the actual sum is less than 3^10 and more than 3^7 (option A) as the last term is already more than that. So the answer is clearly B.

We have the sum of 9 terms. Now, if all terms were equal to the largest term 2*3^7 we would have: sum=9*(2*3^7)=2*3^9=~3^10, so the actual sum is less than 3^10 and more than 3^7 (option A) as the last term is already more than that. So the answer is clearly B.

We have the sum of 9 terms. Now, if all terms were equal to the largest term 2*3^7 we would have: sum=9*(2*3^7)=2*3^9=~3^10, so the actual sum is less than 3^10 and more than 3^7 (option A) as the last term is already more than that. So the answer is clearly B.

Hi Karishma, Could you please explain the last step. How did you get 3^8 and divided by (2-1)

There are two ways to get that: 1. The number property discussed above: \(2( 3^0 + 3^1 + 3^2 + .... + 3^{n - 1}) = 3^n - 1\) If n = 8, \(2( 3^0 + 3^1 + 3^2 + .... + 3^7) = 3^8 - 1\) \(3^0 + 3^1 + 3^2 + .... + 3^7 = (3^8 - 1)/2\) Since 3^0 = 1, \(1 + 3^1 + 3^2 + .... + 3^7 = (3^8 - 1)/2\) \(3^1 + 3^2 + .... + 3^7 = (3^8 - 1)/2 - 1\) That's how we substituted \((3^8 - 1)/2 - 1\) for \(3^1 + 3^2 + .... + 3^7\)

Another method is using the GP formula \(3^1 + 3^2 + .... + 3^7 = 3(3^7 - 1)/(3 - 1) = (3^8 - 3)/2 = (3^8 - 1 - 2)/2 = (3^8 - 1)/2 - 1\) Here is a post explaining GPs: http://www.veritasprep.com/blog/2012/04 ... gressions/ _________________

The above can be changed to a geometric progression.

3+3+3+2*3^2+2*3^3+.....................+2*3^6

9 + 2( 3^2+...........+3^6)

Sum for a geometric progression given below.

a+ar+ar^2+................+ar^n is

S= a(r^n-1)/(r-1)

Applying this to our series.

9 + 2 { 9 (3^6 - 1)/(3-1)}

=> 3^8
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