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AP EAMCET Engineering 21 Apr 2019 Shift 1 Paper
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© examsnet.com
Question : 17
Total: 160
Let
a
0
,
a
1
,
a
2
, ...
a
n
∊ R be in an arithmetic progression and let
C
0
,
c
1
,
C
2
, ...
C
n
be the binomial coefficients. Then
n
Σ
k
=
0
a
k
.
C
k
=
1
2
(
a
0
+
a
n
)
(
a
0
+
a
n
)
.2
n
−
1
(
a
0
+
a
n
)
0
Validate
Solution:
Consider the expression,
n
∑
k
=
0
a
k
⋅
C
k
=
a
0
C
0
+
a
1
C
1
+
a
2
C
2
+
.
.
.
+
a
n
C
n
=
a
0
C
0
+
(
a
0
+
d
)
C
1
+
(
a
0
+
2
d
)
C
2
+
.
.
.
+
(
a
0
+
n
d
)
C
n
Here
d
is an common difference then
n
∑
k
=
0
a
k
⋅
C
k
=
a
0
(
C
0
+
C
1
+
.
.
.
+
C
n
)
+
d
(
C
1
+
2
C
2
+
3
C
3
+
.
.
.
+
n
C
n
)
=
a
0
2
n
+
d
(
n
2
n
−
1
)
=
2
n
−
1
[
2
a
0
+
n
d
]
=
(
a
0
+
a
n
)
2
n
−
1
© examsnet.com
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