If the set R = { ( a , b ) : a + 5 b = 42 , ? a , b ? N } has m elements and ? n = 1 m ( 1 - i n ! ) = x + i y , ? where i = - 1 , then the value of m + x + y is [2024]

Question

If the set $R = {(a, b) : a + 5 b = 42, a, b \in N}$ has $m$ elements and $\sum_{n = 1}^{m} (1 - i^{n!}) = x + i y,$ where $i = \sqrt{- 1},$ then the value of $m + x + y$ is [2024]

Smart Achievers · Accepted Answer

(4)

Given $R = {(a, b) : a + 5 b = 42, a, b \in N}$

So, $\sum_{n = 1}^{m} (1 - i^{n!}) = x + i y,$

$∴ (a, b) = {(37, 1), (32, 2), (27, 3), (22, 4), (17, 5), (12, 6), (7, 7), (2, 8)}$

$∴ | R | = 8 \Rightarrow m = 8$

Now, $\sum_{n = 1}^{m} (1 - i^{n!}) = x + i y$

Clearly for $n = 4, i^{n!} = i^{4!} = ({{(i)}^{4})}^{6} = {(1)}^{6} = 1$

$∴$ $i^{n!} = 1$ for $n \geq 4$

$∴$ $\sum_{n = 1}^{8} (1 - i^{n!}) = (1 - i) + (1 - i^{2!}) + (1 - i^{3!})$

$= 1 - i + 1 + 1 + 1 + 1 = 5 - i \Rightarrow x = 5 and y = - 1$ .

So, $m + x + y = 8 + 5 - 1 = 12$

Solution

Q.
If the set $R = {(a, b) : a + 5 b = 42, a, b \in N}$ has $m$ elements and $\sum_{n = 1}^{m} (1 - i^{n!}) = x + i y,$ where $i = \sqrt{- 1},$ then the value of $m + x + y$ is [2024]

1 8

2 5

3 4

4 12

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Solution

Q. If the set R={(a,b):a+5b=42, a,b∈N} has m elements and ∑n=1m(1-in!)=x+iy, where i=-1, then the value of m+x+y is [2024]

1 8

2 5

3 4

4 12

Want Us to Email you About Special Offers & Updates?

Q.
If the set $R = {(a, b) : a + 5 b = 42, a, b \in N}$ has $m$ elements and $\sum_{n = 1}^{m} (1 - i^{n!}) = x + i y,$ where $i = \sqrt{- 1},$ then the value of $m + x + y$ is [2024]