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$\begin{array}{}Let\;f_{n+1}(x)&=f_n(x)+1\;\text{if n is a multiple of 3}\\ &=f_n(x)-1\;\text{otherwise.}\end{array}$

If $f_1(1)=0$, then what is $f_{50}(1)$?

1. $-18$
2. $-16$
3. $-17$
4. Cannot be determined

edited | 38 views

Given that $:\begin{array}{}\;f_{n+1}(x)&=f_n(x)+1\;\text{if n is a multiple of 3}\\ &=f_n(x)-1\;\text{otherwise.}\end{array}$

and  $n = 0,f_1(1)=0$

Now $n = 1,f_{2}(1) = f_{1}(1) -1 = 0 -1 = -1$

$n = 2,f_{3}(1) = f_{2}(1) -1 = -1-1 = -2$

$n = 3,f_{4}(1) = f_{3}(1) + 1 = -2 + 1 = -1$

$n = 4,f_{5}(1) = f_{4}(1) -1 = -1-1 = -2$

$n = 5,f_{6}(1) = f_{5}(1) -1 = -2-1 = -3$

$n = 6,f_{7}(1) = f_{6}(1) +1 = -3+1 = -2$

$n = 7,f_{8}(1) = f_{7}(1) -1 = -2-1 = -3$

$n = 8,f_{9}(1) = f_{8}(1) -1 = -3-1 = -4$

$n = 9,f_{10}(1) = f_{9}(1) +1 = -4+1 = -3$

Now, we got the pattern , every multiple of $’3’$, the number decreases by $’-1’.$

• $n=3,f_{4}(1)\implies -1$
• $n=6,f_{7}(1)\implies -2$
• $n=9,f_{10}(1)\implies -3$
• $n=12,f_{13}(1)\implies -4$
• $n=15,f_{16}(1)\implies -5$
• $n=18,f_{19}(1)\implies -6$
• $n=21,f_{22}(1)\implies -7$
• $n=24,f_{25}(1)\implies -8$
• $n=27,f_{28}(1)\implies -9$
• $n=30,f_{31}(1)\implies -10$
• $n=33,f_{34}(1)\implies -11$
• $n=36,f_{37}(1)\implies -12$
• $n=39,f_{40}(1)\implies -13$
• $n=42,f_{43}(1)\implies -14$
• $n=45,f_{46}(1)\implies -15$
• $n = 48,f_{49}(1)\implies -16$
• $n = 49,f_{50}(1) = f_{49}(1)-1 = -16-1 = -17.$

$$\text{(OR)}$$

$3\implies -1\Leftrightarrow 48 \implies -16$

Now, $n = 49,f_{50}(1) = f_{49}(1)-1 = -16-1 = -17.$

So, the correct answer is $(C).$

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