Compute i^1+i^2+i^3....i^99+i^100
1 answer:
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Answer:
<h3>i¹ + i² + i³ +. . .+ i⁹⁹ + i¹⁰⁰ = 0</h3>Step-by-step explanation:
Consider the sum S = i¹ + i² + i³ +. . .+ i⁹⁹ + i¹⁰⁰
S = i¹ + i² + i³ + . . . + i⁹⁹ + i¹⁰⁰
S = a₁ + a₂ + a₃ +. . . + a₉₉ + a₁₀₀
then, S is the sum of 100 consecutive terms of a geometric sequence (an)
where the first term a1 = i¹ = i and the common ratio = i
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then
or i¹⁰⁰ = (i⁴)²⁵ = 1²⁵ = 1 (we know that i⁴ = 1)
Hence
S = 0
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<h2>Answer:</h2>
Ques 1)
Ques 2)
Ques 1)
We know that if a graph is stretched by a factor of a then the transformation if given by:
f(x) → a f(x)
Also, we know that the translation of a function k units to the right or to the left is given by:
f(x) → f(x+k)
where if k>0 then the shift is k units to the left
and if k<0 then the shift is k units to the right.
Here the graph of f(x) is transformed into the graph of g(x) by a vertical stretch of 4 units and a translation of 4 units right.
This means that the function g(x) is given by:
Ques 2)
We know that the transformation of the type:
f(x) → f(x)+k
is a shift or translation of the function k units up or down depending on k.
If k>0 then the shift is k units up.
and if k<0 then the shift is k units down.
Here, The graph of the function f(x)=|3x| is translated 4 units up.
This means that the transformed function g(x) is given by: