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Joint Entrance Examination
Mathematics
Binomial Theorem
Question 84
Question 84 of 347
Medium
NUMERICAL
11
C
1
2
+
11
C
2
3
+
…
+
11
C
9
10
=
n
m
g
c
d
(
n
,
m
)
=
1
n
+
m
\frac{{}^{11}{C}_{1}}{2}+\frac{{}^{11}{C}_{2}}{3}+\ldots +\frac{{}^{11}{C}_{9}}{10}=\frac{n}{m}gcd(n,m)=1n+m
2
11
C
1
+
3
11
C
2
+
…
+
10
11
C
9
=
m
n
g
c
d
(
n
,
m
)
=
1
n
+
m
\_\_\_\_\_\_\_. Answer: 2041
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Official solution
∑
r
=
1
9
11
C
r
r
+
1
=
1
12
∑
r
=
1
9
12
C
r
+
1
=
1
12
[
2
12
−
26
]
=
2035
6
∴
m
+
n
=
2041
\begin{gathered} \sum _{r=1}^{9}\frac{{}^{11}{C}_{r}}{r+1} \\ =\frac{1}{12}\sum _{r=1}^{9}{}^{12}{C}_{r+1} \\ =\frac{1}{12}[{2}^{12}-26]=\frac{2035}{6} \\ \therefore m+n=2041 \end{gathered}
r
=
1
∑
9
r
+
1
11
C
r
=
12
1
r
=
1
∑
9
12
C
r
+
1
=
12
1
[
2
12
−
26
]
=
6
2035
∴
m
+
n
=
2041
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