Question

) Find the value of 1 + i^2 + i^4 +i6+i^8+...+i^20​

Answer 1

Answer:

1

Step-by-step explanation:

The given sequence is geometric progression

where

$first \: term, a = 1 \\ </p><p>Last \: term, T_n = {i}^{20} \\ </p><p>Common \: ration, r = a_2/a_1 = i^2 /1 = i^2$

To find n

$T_n = a {r}^{n - 1} \\ {i}^{20} = 1 \times {( {i}^{2}) }^{n - 1} \\ { ({i}^{2}) }^{10} = { ({i}^{2} )}^{n - 1} \\ as \: both \: the \: terms \: are \: equal \: hence \\ n - 1 = 10 \\ n = 11$

Sum of GP is

$S_n = \frac{a( {r}^{n} - 1) }{r - 1} \\ = \frac{1( {( { i }^{2}) }^{n} - 1) }{ {i}^{2} - 1} \\ = \frac{( {( { i }^{2}) }^{11} - 1) }{ {i}^{2} - 1} \\ = \frac{ { i }^{22} - 1 }{ {i}^{2} - 1} = 1$

Answer 2

Answer:

Kindly see the Attachments (2)

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