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Byju's Answer
Standard XII
Mathematics
Sum of Coefficients of All Terms
The value of ...
Question
The value of
1
+
i
2
+
i
4
+
i
6
+
.
.
.
.
+
i
2
n
is
A
Positive
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B
Negative
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C
Zero
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D
Cannot be determined
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Solution
The correct option is
B
Cannot be determined
i
1
=
i
i
2
=
−
1
i
3
=
−
i
i
4
=
1
Other powers of i can be easily found out.
i
5
=
i
2
×
i
3
=
−
1
×
−
i
=
i
Similarly
i
6
=
i
4
×
i
2
=
1
×
−
1
=
−
1
i
2
n
=
−
1
when n is odd that is n=1,3,5...
i
2
n
=
1
when n is even that is n=2,4,6...
In the question
if n is odd the last term will be -1 and hence the summation is:
1
+
i
2
+
i
4
+
i
6
+
.
.
.
.
+
i
2
n
=
1
+
(
−
1
)
+
1
(
−
1
)
+
1......
(
−
1
)
=
0
if n is even the last term will be 1 and hence the summation is:
1
+
i
2
+
i
4
+
i
6
+
.
.
.
.
+
i
2
n
=
1
+
(
−
1
)
+
1
(
−
1
)
+
1......1
=
1
So the answer is
0
or
1
depending on whether n is odd or even.
Unless
n
is specified exact answer cannot be determined.
Suggest Corrections
0
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