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Byju's Answer
Standard IX
Mathematics
Laws of Exponents for Real Numbers
Solve 2n+4-2...
Question
Solve
2
n
+
4
−
2.2
n
2.2
n
+
3
+
2
−
3
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Solution
Given,
2
n
+
4
−
2
⋅
2
n
2.2
n
+
3
+
2
−
3
=
2
n
⋅
2
4
−
2
⋅
2
n
2.2
n
⋅
2
3
+
2
−
3
=
2
n
(
2
4
−
2
)
2.2
n
⋅
2
3
+
2
−
3
=
2
4
−
2
2
⋅
2
3
+
2
−
3
=
14
16
+
2
−
3
=
7
8
+
1
8
=
7
+
1
8
=
8
8
=
1
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0
Similar questions
Q.
Evaluate :
2
n
+
4
−
2.2
n
2.2
n
+
3
+
2
−
3
Q.
Evaluate:
(
i
)
16.
2
n
+
1
−
4.
2
n
16.
2
n
+
2
−
2.
2
n
+
2
(
i
i
)
1
1
+
√
2
+
1
√
2
+
√
3
+
1
√
3
+
√
4
Q.
If
C
0
,
C
1
,
C
2
,
.
.
.
.
.
C
n
denote the coefficient in the expansion of
(
1
+
x
)
n
, then the value of
C
1
+
2
C
2
+
3
C
3
.
.
.
.
+
n
C
n
is
Q.
Solve :
lim
n
→
∞
1
+
3
+
5
+
…
.
.
.
+
.
(
2
n
−
1
)
2
+
4
+
6
+
.
.
+
2
n
Q.
Column - I
Column - II
(A)
∑
n
r
=
1
r
.
n
C
r
is equal to
(P)
(
n
+
2
)
.2
n
−
1
(B)
∑
n
+
1
r
=
1
r
.
n
C
r
−
1
is equal to
(Q)
(
n
+
1
)
.
2
n
C
n
(C)
∑
n
r
=
0
(
2
r
+
1
)
.
n
C
r
is equal to
(R)
(
n
+
1
)
.2
n
(D)
∑
n
r
=
0
(
2
r
+
1
)
.
(
n
C
r
)
2
is equal to
(S)
n
.2
n
−
1
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