1눀+2눀+3눀+4눀+……+n눀=?

2024-10-31 04:03:01
推荐回答(5个)
回答1:

1²+2²+3²+4²+……+n²=n(n+1)(2n+1)/6

1²+2²+3²+4²+……+n²

=1*(2-1)+……n*(n+1-1)

=1*2+2*3+……+n*(n+1)-(1+2+……+n)

=2*(2C1+3C2+……+(n+1)Cn)(C为排列组合标志)-n*(n+1)/2

=(n+2)C3+1-n*(n+1)/2

=n(n+1)(2n+1)/6

扩展资料:

相关公式:

(1)1+2+3+.+n=n(n+1)/2

(2)1^2+2^2+3^2+...+n^2=n(n+1)(2n+1)/6

(3)1×2+2×3+3×4+4×5+…+n(n+1)

=(1^2+1)+(2^2+2)+(3^2+2)+...+(n^2+n)

=(1^2+2^2+...+n^2)+(1+2+3+.+n)

=n(n+1)(2n+1)/6+n(n+1)/2

=n(n+1)(n+2)

(4)1×2×3+2×3×4+3×4×5+......+n(n+1)(n+2)

=1/4【1×2×3×4-0×1×2×3】+1/4【2×3×4×5-1×2×3×4】+1/4【3×4×5×6-2×3×4×5】+......+

1/4【n(n+1)(n+2)(n+3)-(n-1)n(n+1)(n+2)】

=1/4×n(n+1)(n+2)(n+3)

回答2:

1²+2²+3²+4²+……+n²=n(n+1)(2n+1)/6

1²+2²+3²+4²+……+n²

=1*(2-1)+……n*(n+1-1)

=1*2+2*3+……+n*(n+1)-(1+2+……+n)

=2*(2C1+3C2+……+(n+1)Cn)(C为排列标志)-n*(n+1)/2

=(n+2)C3+1-n*(n+1)/2

=n(n+1)(2n+1)/6

扩展资料:

类似的公式:

1+2+...+n=n(n+1)/2;

1³+2³+...+n³=[n(n+1)/2]²;

1+3+5+...+(2n-1)=n²。

常用平方数:

1² = 1, 2² = 4 ,3² = 9, 4² = 16, 5² = 25, 6² = 36 ,7² = 49 ,8² = 64 ,9² = 81 ,10² = 100。

11² = 121, 12² = 144 ,13² = 169 ,14² = 196 ,15² = 225, 16² = 256, 17² = 289 ,18² = 324, 19² = 361 ,20² = 400。

21² = 441 ,22² = 484, 23² = 529 ,24² = 576, 25² = 625 ,26² = 676, 27² = 729 ,28² = 784 ,29² = 841, 30² = 900。

回答3:

S1(n)=1+2+3+…+n,
S2(n)=12+22+32+…+n2,…
Sk(n)=1k+2k+3k+…+nk (k∈N*).
已知
13= 1,
23=(1+1)3=13+3×12+3×1+1,
33=(2+1)3=23+3×22+3×2+1,
43=(3+1)3=33+3×32+3×3+1,……
n3=(n-1)3+3(n-1)2+3(n-1)+1.
将左右两边分别相加,得
S3(n)=[S3(n)-n3]+3[S2(n)-n2]+3[S1(n)-n]+n.
由此知
S2(n)=n3+3n2+2n-3S1(n)3=2n3+3n2+n6
=n(n+1)(2n+1)6.

回答4:


参考

回答5:

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