The value of t2-t4-t6-t20 if tn-in- i-1 is

Given that nbsp;t_n=i^n; i=√( 1) To find nbsp;t_2+t_4+t_6+...+t_20 ⇒ i^2+i^4+i^6+...+i^20 ⇒ i^2+i^2.2+i^2.3+...+i^2.10 ⇒ 1+( 1)^2+( 1)^3+( 1)^4+...+( 1)^10 ...

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The value of ...

Question

The value of $t_{2} + t_{4} + t_{6} + . . . + t_{20}$ if $t_{n} = i^{n}; i = \sqrt{- 1}$ is

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Solution

Given that $t_{n} = i^{n}; i = \sqrt{- 1}$
To find $t_{2} + t_{4} + t_{6} + . . . + t_{20}$
$\Rightarrow i^{2} + i^{4} + i^{6} + . . . + i^{20}$
$\Rightarrow i^{2} + i^{2.2} + i^{2.3} + . . . + i^{2.10}$
$\Rightarrow - 1 + (- 1)^{2} + (- 1)^{3} + (- 1)^{4} + . . . + (- 1)^{10}$
$\Rightarrow - 1 + 1 - 1 + 1.... - 1 + 1 = 0$

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Similar questions

Q. In an A.P. if

S_{1} = T_{1} + T_{2} + T_{3} + . . . + T_{n}

(n odd),

S_{2} = T_{2} + T_{4} + T_{6} + . . . + T_{n - 1}

, then find the value of

S_{1} / S_{2}

in terms of n

Q. In an A.P., if

S_{1} = T_{1} + T_{2} + T_{3} + . . . . . . + T_{n}

(

n

odd) and

S_{2} = T_{2} + T_{4} + T_{6} + . . . . + T_{n - 1}

, then find the value of

\frac{S_{1}}{S_{2}}

in terms of

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be terms of an A.P. If

S_{1} = T_{1} + T_{2} + T_{3} + \dots + T_{n}

and

S_{2} = T_{2} + T_{4} + T_{6} + \dots + T_{n - 1}

, where

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\frac{S_{1}}{S_{2}}

Q. If

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represent the term in the expansion of

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Q. If

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represent the terms in the expansion of

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, then the value of

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{(T_{1} - T_{3} + T_{5} - . . . .)}_{1}^{2}

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The value of t2+t4+t6+...+t20 if tn=in; i=√−1 is

Given that tn=in; i=√−1 To find t2+t4+t6+...+t20 ⇒i2+i4+i6+...+i20 ⇒i2+i2.2+i2.3+...+i2.10 ⇒−1+(−1)2+(−1)3+(−1)4+...+(−1)10 ⇒−1+1−1+1....−1+1=0

The value of $t_{2} + t_{4} + t_{6} + . . . + t_{20}$ if $t_{n} = i^{n}; i = \sqrt{- 1}$ is