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Manipulations de sommes et de produits
Sujet
2
2
2
- Exercice 1
10 min
25
Question 1
Soit
n
n
n
un entier naturel. Calculer :
S
1
=
∑
(
i
,
j
)
∈
[
[
1
,
n
]
]
2
m
i
n
(
i
,
j
)
S_1=\sum_{\left(i,j\right)\in {\left[\left[1,n\right]\right]}^2}{{\mathrm{min} \left(i,j\right)\ }}
S
1
=
(
i
,
j
)
∈
[
[
1
,
n
]
]
2
∑
min
(
i
,
j
)
Correction
S
1
=
∑
(
i
,
j
)
∈
[
[
1
,
n
]
]
2
m
i
n
(
i
,
j
)
S_1=\sum_{\left(i,j\right)\in {\left[\left[1,n\right]\right]}^2}{{\mathrm{min} \left(i,j\right)\ }}
S
1
=
(
i
,
j
)
∈
[
[
1
,
n
]
]
2
∑
min
(
i
,
j
)
équivaut successivement à :
S
1
=
∑
i
=
1
n
∑
j
=
1
n
m
i
n
(
i
,
j
)
S_1=\sum^n_{i=1}{\sum^n_{j=1}{{\mathrm{min} \left(i,j\right)\ }}}
S
1
=
i
=
1
∑
n
j
=
1
∑
n
min
(
i
,
j
)
S
1
=
∑
i
=
1
n
(
∑
j
=
1
i
m
i
n
(
i
,
j
)
+
∑
j
=
i
+
1
n
m
i
n
(
i
,
j
)
)
S_1=\sum^n_{i=1}{\left(\sum^i_{j=1}{{\mathrm{min} \left(i,j\right)\ }}+\sum^n_{j=i+1}{{\mathrm{min} \left(i,j\right)\ }}\right)}
S
1
=
i
=
1
∑
n
(
j
=
1
∑
i
min
(
i
,
j
)
+
j
=
i
+
1
∑
n
min
(
i
,
j
)
)
S
1
=
∑
i
=
1
n
(
∑
j
=
1
i
j
+
∑
j
=
i
+
1
n
i
)
S_1=\sum^n_{i=1}{\left(\sum^i_{j=1}{j}+\sum^n_{j=i+1}{i}\right)}
S
1
=
i
=
1
∑
n
(
j
=
1
∑
i
j
+
j
=
i
+
1
∑
n
i
)
Soit
α
\alpha
α
un réel . Pour tout entier naturel
n
n
n
, on a :
\;
∑
k
=
1
n
α
=
n
α
\sum^n_{k=1}{\alpha=n\alpha}
k
=
1
∑
n
α
=
n
α
\;
∑
k
=
1
n
k
=
n
(
n
+
1
)
2
\sum^n_{k=1}{k=\frac{n\left(n+1\right)}{2}}
k
=
1
∑
n
k
=
2
n
(
n
+
1
)
S
1
=
∑
i
=
1
n
(
i
(
i
+
1
)
2
+
(
n
−
(
i
+
1
)
+
1
)
×
i
)
S_1=\sum^n_{i=1}{\left(\frac{i\left(i+1\right)}{2}+\left(n-\left(i+1\right)+1\right)\times i\right)}
S
1
=
i
=
1
∑
n
(
2
i
(
i
+
1
)
+
(
n
−
(
i
+
1
)
+
1
)
×
i
)
S
1
=
∑
i
=
1
n
(
i
(
i
+
1
)
2
+
(
n
−
i
)
×
i
)
S_1=\sum^n_{i=1}{\left(\frac{i\left(i+1\right)}{2}+\left(n-i\right)\times i\right)}
S
1
=
i
=
1
∑
n
(
2
i
(
i
+
1
)
+
(
n
−
i
)
×
i
)
S
1
=
∑
i
=
1
n
(
1
2
i
2
+
1
2
i
+
n
i
−
i
2
)
S_1=\sum^n_{i=1}{\left(\frac{1}{2}i^2+\frac{1}{2}i+ni-i^2\right)}
S
1
=
i
=
1
∑
n
(
2
1
i
2
+
2
1
i
+
ni
−
i
2
)
S
1
=
∑
i
=
1
n
(
−
1
2
i
2
+
1
2
i
+
n
i
)
S_1=\sum^n_{i=1}{\left(-\frac{1}{2}i^2+\frac{1}{2}i+ni\right)}
S
1
=
i
=
1
∑
n
(
−
2
1
i
2
+
2
1
i
+
ni
)
S
1
=
∑
i
=
1
n
(
−
1
2
i
2
+
(
1
2
+
n
)
i
)
S_1=\sum^n_{i=1}{\left(-\frac{1}{2}i^2+\left(\frac{1}{2}+n\right)i\right)}
S
1
=
i
=
1
∑
n
(
−
2
1
i
2
+
(
2
1
+
n
)
i
)
S
1
=
∑
i
=
1
n
(
−
1
2
i
2
)
+
∑
i
=
1
n
(
(
1
2
+
n
)
i
)
S_1=\sum^n_{i=1}{\left(-\frac{1}{2}i^2\right)}+\sum^n_{i=1}{\left(\left(\frac{1}{2}+n\right)i\right)}
S
1
=
i
=
1
∑
n
(
−
2
1
i
2
)
+
i
=
1
∑
n
(
(
2
1
+
n
)
i
)
S
1
=
−
1
2
∑
i
=
1
n
(
i
2
)
+
(
1
2
+
n
)
∑
i
=
1
n
(
i
)
S_1=-\frac{1}{2}\sum^n_{i=1}{\left(i^2\right)}+\left(\frac{1}{2}+n\right)\sum^n_{i=1}{\left(i\right)}
S
1
=
−
2
1
i
=
1
∑
n
(
i
2
)
+
(
2
1
+
n
)
i
=
1
∑
n
(
i
)
Pour tout entier naturel
n
n
n
, on a :
\;
∑
k
=
1
n
k
=
n
(
n
+
1
)
2
\sum^n_{k=1}{k=\frac{n\left(n+1\right)}{2}}
k
=
1
∑
n
k
=
2
n
(
n
+
1
)
\;
∑
k
=
1
n
k
2
=
n
(
n
+
1
)
(
2
n
+
1
)
6
\sum^n_{k=1}{k^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6}}
k
=
1
∑
n
k
2
=
6
n
(
n
+
1
)
(
2
n
+
1
)
S
1
=
−
1
2
×
n
(
n
+
1
)
(
2
n
+
1
)
6
+
(
1
2
+
n
)
×
n
(
n
+
1
)
2
S_1=-\frac{1}{2}\times \frac{n\left(n+1\right)\left(2n+1\right)}{6}+\left(\frac{1}{2}+n\right)\times \frac{n\left(n+1\right)}{2}
S
1
=
−
2
1
×
6
n
(
n
+
1
)
(
2
n
+
1
)
+
(
2
1
+
n
)
×
2
n
(
n
+
1
)
S
1
=
−
n
(
n
+
1
)
(
2
n
+
1
)
12
+
(
1
2
+
n
)
n
(
n
+
1
)
2
S_1=\frac{-n\left(n+1\right)\left(2n+1\right)}{12}+\frac{\left(\frac{1}{2}+n\right)n\left(n+1\right)}{2}
S
1
=
12
−
n
(
n
+
1
)
(
2
n
+
1
)
+
2
(
2
1
+
n
)
n
(
n
+
1
)
S
1
=
(
n
(
n
+
1
)
)
(
−
(
2
n
+
1
)
12
+
(
1
2
+
n
)
2
)
S_1=\left(n\left(n+1\right)\right)\left(\frac{-\left(2n+1\right)}{12}+\frac{\left(\frac{1}{2}+n\right)}{2}\right)
S
1
=
(
n
(
n
+
1
)
)
(
12
−
(
2
n
+
1
)
+
2
(
2
1
+
n
)
)
S
1
=
(
n
(
n
+
1
)
)
(
−
2
n
−
1
12
+
3
+
6
n
12
)
S_1=\left(n\left(n+1\right)\right)\left(\frac{-2n-1}{12}+\frac{3+6n}{12}\right)
S
1
=
(
n
(
n
+
1
)
)
(
12
−
2
n
−
1
+
12
3
+
6
n
)
S
1
=
(
n
(
n
+
1
)
)
(
−
2
n
−
1
+
3
+
6
n
12
)
S_1=\left(n\left(n+1\right)\right)\left(\frac{-2n-1+3+6n}{12}\right)
S
1
=
(
n
(
n
+
1
)
)
(
12
−
2
n
−
1
+
3
+
6
n
)
S
1
=
(
n
(
n
+
1
)
)
(
4
n
+
2
12
)
S_1=\left(n\left(n+1\right)\right)\left(\frac{4n+2}{12}\right)
S
1
=
(
n
(
n
+
1
)
)
(
12
4
n
+
2
)
Ainsi :
S
1
=
(
n
(
n
+
1
)
)
(
2
n
+
1
6
)
S_1=\left(n\left(n+1\right)\right)\left(\frac{2n+1}{6}\right)
S
1
=
(
n
(
n
+
1
)
)
(
6
2
n
+
1
)