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4-1.Complex numbers
medium
The value of ${i^{1 + 3 + 5 + ... + (2n + 1)}}$ is
A
$i$ if $n$ is even, $-i$ if $n$ is odd
B
$1$ if $n$ is even, $-1$ if $n$ is odd
C
$1$ if $n$ is odd, $-1$ if $n$ is even
D
$i$ if $n$ is even,$ -1$ if $n$ is odd
Solution
(c) Let $z = i^{[1 + 3 + 5 + …. + (2n + 1)]}$
Clearly series is A.P. with common difference $= 2$
$\because \,T_n = 2n – 1$and ${T_{n + 1}} = 2n + 1$
So, number of terms in A. P. $ = n + 1$
Now, ${S_{n + 1}} = \frac{{n + 1}}{2}[2.1 + (n + 1 – 1)2]$
$ \Rightarrow {S_{n + 1}} = \frac{{n + 1}}{2}[2 + 2n] = (n + 1)^2$ i.e. $i^{(n + 1)^2}$
Now put $n = 1,\,2,\,3,\,4,\,5,\,…..$
$n = 1,z = {i^4} = 1$, $n = 2,\,z = {i^6} = – 1$,
$n = 3,\,z = {i^8} = 1$, $n = 4,\,z = {i^{10}} = – 1$,
$n = 5,\,\,z = {i^{12}} = 1\,,……..$
Standard 11
Mathematics
