Bunuel
Competition Mode Question
If \((10)^9 + 2(11)^1(10)^8 + 3(11)^2(10)^7 + … + 10(11)^9 = k(10)^9\), then k is equal to
A. 121/10
B. 441/100
C. 100
D. 110
E. 120
Since 11 = 1.1 x 10, we can rewrite the left hand side of the equation as:
10^9 + 2(1.1 x 10)(10^8) + 3(1.1^2 x 10^2)(10^7) + … + 10(1.1^9 x 10^9) = k(10^9)
10^9 + 2(1.1)(10^9) + 3(1.1^2)(10^9) + … + 10(1.1^9 )(10^9) = k(10^9)
Dividing the equation by 10^9, we have:
1 + 2(1.1) + 3(1.1)^2 + … + 10(1.1^9) = k → Eq. 1
We need to find the value of k. Without a calculator, we can multiply the equation by 1.1 to obtain:
1.1 + 2(1.1)^2 + 3(1.1)^3 + … + 10(1.1^10) = 1.1k → Eq. 2
Subtracting Eq. 2 from Eq. 1 (by combining terms with the same powers of 1.1), we have:
1 + 1.1 + 1.1^2 + 1.1^3 + … + 1.1^9 - 10(1.1^10) = -0.1k → Eq. 3
Now we see that all the terms (except the last term) form a geometric sequence (notice that the first term 1 can be written as 1.1^0).
Recall that the sum of a finite geometric series is:
S_n = a_1 * (1 - r^n) / (1 - r)
where a_1 is the first term, n is the number of terms and r is the common ratio.
So the sum of the 10 positive terms (i.e., 1, 1.1, 1.1^2, …, 1.1^9) is:
S_10 = 1(1 - 1.1^10)/(1 - 1.1) = (1 - 1.1^10)/(-0.1)
Substituting this back into Eq. 3, we have:
(1 - 1.1^10)/(-0.1) - 10(1.1^10) = -0.1k
Multiply the above equation by -0.1, we have:
1 - 1.1^10 + 1.1^10 = 0.01k
1 = 0.01k
100 = k
Answer: C