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Binomial Theorem
NDA 2017
Exam
NDA
Year
2017
Subject
Mathematics
Chapter
Binomial Theorem
NDA 2017 · Mathematics · Binomial Theorem
The value of [C(7,0)+C(7,1)]+[C(7,1)+C(7,2)]+ ..+[C(7,6)+C(7,7)] is
The value of
[
C
(
7
,
0
)
+
C
(
7
,
1
)
]
+
[
C
(
7
,
1
)
+
C
(
7
,
2
)
]
+
…
.
.
+
[
C
(
7
,
6
)
+
C
(
7
,
7
)
]
[\mathrm{C}(7,0)+\mathrm{C}(7,1)]+[\mathrm{C}(7,1)+\mathrm{C}(7,2)]+\ldots..+[\mathrm{C}(7,6)+\mathrm{C}(7,7)]
[
C
(
7
,
0
)
+
C
(
7
,
1
)]
+
[
C
(
7
,
1
)
+
C
(
7
,
2
)]
+
…
..
+
[
C
(
7
,
6
)
+
C
(
7
,
7
)]
is
Options
A
254
B
255
C
256
D
257
Correct answer
A. 254
Step-by-step solution
(
7
c
0
+
7
c
1
)
+
\left(7_{c_0}+7_{c_1}\right)+
(
7
c
0
+
7
c
1
)
+
\left(7_{c_1}+7_{c_2}\right)+\ldots .+
(
7
c
6
+
7
c
7
)
\left(7_{c_6}+7_{c_7}\right)
(
7
c
6
+
7
c
7
)
We know,
n
c
r
+
n
c
r
−
1
=
n
+
1
c
r
\mathrm{n}_{\mathrm{c}_{\mathrm{r}}}+\mathrm{n}_{\mathrm{c}_{\mathrm{r}-1}}=\mathrm{n}+1_{\mathrm{c}_{\mathrm{r}}}
n
c
r
+
n
c
r
−
1
=
n
+
1
c
r
=
8
c
1
+
8
c
2
+
…
.
.
+
8
c
7
=
(
8
c
0
+
8
c
1
+
8
c
2
+
…
.
.
+
8
c
7
+
8
c
8
)
−
(
8
c
0
+
8
c
8
)
=
2
8
−
(
1
+
1
)
[
Since,
n
c
0
+
n
c
1
+
n
c
2
+
…
.
+
n
c
n
=
2
n
]
=
256
−
2
=
254
\begin{aligned} & =8_{\mathrm{c}_1}+8_{\mathrm{c}_2}+\ldots . .+8_{\mathrm{c}_7} \\ & =\left(8_{\mathrm{c}_0}+8_{\mathrm{c}_1}+8_{\mathrm{c}_2}+\ldots . .+8_{\mathrm{c}_7}+8_{\mathrm{c}_8}\right)-\left(8_{\mathrm{c}_0}+8_{\mathrm{c}_8}\right) \\ & =2^8-(1+1) \\ & {\left[\text{ Since, } \mathrm{n}_{\mathrm{c}_0}+\mathrm{n}_{\mathrm{c}_1}+\mathrm{n}_{\mathrm{c}_2}+\ldots .+\mathrm{n}_{\mathrm{c}_{\mathrm{n}}}=2^{\mathrm{n}}\right]} \\ & =256-2 \\ & =254 \end{aligned}
=
8
c
1
+
8
c
2
+
…
..
+
8
c
7
=
(
8
c
0
+
8
c
1
+
8
c
2
+
…
..
+
8
c
7
+
8
c
8
)
−
(
8
c
0
+
8
c
8
)
=
2
8
−
(
1
+
1
)
[
Since,
n
c
0
+
n
c
1
+
n
c
2
+
…
.
+
n
c
n
=
2
n
]
=
256
−
2
=
254
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