MathDB
Problems
Contests
National and Regional Contests
Japan Contests
Today's Calculation Of Integral
2007 Today's Calculation Of Integral
199
199
Part of
2007 Today's Calculation Of Integral
Problems
(1)
Today's calculation of Integral 199
Source:
4/23/2007
Let
m
,
n
m,\ n
m
,
n
be non negative integers. Calculate
∑
k
=
0
n
(
−
1
)
k
n
+
m
+
1
k
+
m
+
1
n
C
k
.
\sum_{k=0}^{n}(-1)^{k}\frac{n+m+1}{k+m+1}\ nC_{k}.
k
=
0
∑
n
(
−
1
)
k
k
+
m
+
1
n
+
m
+
1
n
C
k
.
where
i
C
j
_{i}C_{j}
i
C
j
is a binomial coefficient which means
i
⋅
(
i
−
1
)
⋯
(
i
−
j
+
1
)
j
⋅
(
j
−
1
)
⋯
2
⋅
1
\frac{i\cdot (i-1)\cdots(i-j+1)}{j\cdot (j-1)\cdots 2\cdot 1}
j
⋅
(
j
−
1
)
⋯
2
⋅
1
i
⋅
(
i
−
1
)
⋯
(
i
−
j
+
1
)
.
calculus
integration
calculus computations