[GMAT math practice question]
The function is defined as follows: f(x+1) = 3x + 2. What is f(1) + f(2) + f(3) +...+ f(9) + f(10)?
A. 149
B. 153
C. 155
D. 157
E. 159
The function is defined as follows: f(x+1) = 3x + 2. What i
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Since f(x+1) = 3x + 2, we have f(x) = f((x-1) + 1) = 3(x - 1) + 2 = 3x - 3 + 2 = 3x - 1.
f(1) + f(2) + f(3) + … + f(9) + f(10)
= ( 3*1 – 1 ) + ( 3*2 – 1 ) + … + ( 3*10 – 1 )
= 3(1 + 2 + … + 10) – (1 + 1 +…+ 1)
= 3*55 – 10 = 155.
Therefore, C is the answer.
Answer: C
Since f(x+1) = 3x + 2, we have f(x) = f((x-1) + 1) = 3(x - 1) + 2 = 3x - 3 + 2 = 3x - 1.
f(1) + f(2) + f(3) + … + f(9) + f(10)
= ( 3*1 – 1 ) + ( 3*2 – 1 ) + … + ( 3*10 – 1 )
= 3(1 + 2 + … + 10) – (1 + 1 +…+ 1)
= 3*55 – 10 = 155.
Therefore, C is the answer.
Answer: C
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f(x+1) = 3x + 2
Simplify f(x)
f(x+1-1) = 3(x-1) + 2
f(x) = 3x - 3 + 2
The sum of the first f(10) is calculated using
$$S_{\left(n\right)}=\frac{n\left(a+1\right)}{2}$$
Where S(n) = sum of f(10) or S[f(10)]
n = number of terms = 10
a = first term = f(1)
using f(x) = 3x + 1;
f(1) = 3(1) - 1 = 2
f(10) = 3(10) - 1 = 29
$$S_{\left[f\left(10\right)\right]}=\frac{10\left(2+29\right)}{2}=\frac{10\left(31\right)}{2}=\frac{310}{2}=155$$
Answer = option C
Simplify f(x)
f(x+1-1) = 3(x-1) + 2
f(x) = 3x - 3 + 2
The sum of the first f(10) is calculated using
$$S_{\left(n\right)}=\frac{n\left(a+1\right)}{2}$$
Where S(n) = sum of f(10) or S[f(10)]
n = number of terms = 10
a = first term = f(1)
using f(x) = 3x + 1;
f(1) = 3(1) - 1 = 2
f(10) = 3(10) - 1 = 29
$$S_{\left[f\left(10\right)\right]}=\frac{10\left(2+29\right)}{2}=\frac{10\left(31\right)}{2}=\frac{310}{2}=155$$
Answer = option C