= − 𝑘−2 (+2𝑣)
And
Let k = 2
− 0 (2 + 2)2 − 1 (3 + 2)3 − 2 (4 + 2)4
2
=
Let k = 3
3
=
Let k = 4
4
=
− 0 (2 + 2)2
−
→ 4
=
(4 + 2)4
0 2 ∗ 4 ∗ (2 + 2) ∗ (4 + 2)
→ 4
=
− 0 2 ∗ 4 ∗ 6 ∗ (2 + 2) ∗ (4 + 2) ∗ (6 + 2)
6
=
Then according to the pattern
(−1) 0 2 2 ∗ ! ∗ (1 + ) ∗ (2 + ) ∗ (3 + ) ∗ ( + )
2
=
Let k = 5
− 3 (5 + 2)5
5
=
− 1 (3 + 2)3
−
5
=
(5 + 2)5
− 1 3 ∗ 5 ∗ (3 + 2) ∗ (5 + 2)
5
=
According to the pattern 2+1 =
(−1) 1 1 ∗ 3 ∗ 5 ∗ … ∗ (2 + 1) ∗ (1 + ) ∗ (2 + ) ∗ (3 + ) ∗ … ∗ (2 + 1 + ) Let 1 = 0 So 2+1 = 0 Define 0 = 1 2 𝑣 ∗ Γ(1 + v) Where Γ(1 + a) = aΓ(a)
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