蔓芷清芬
幼苗
共回答了24个问题采纳率:91.7% 举报
统一把他们的和记为Sn
1)Sn=1+2+3+.+n
=n+(n-1)+(n-2)+...+1
上下两个配对,为n个n+1,相加得 2Sn=n(n+1)
所以Sn=n(n+1)/2
2)n^2=n(n+1)-n
1^2+2^2+3^2+.+n^2
=1*2-1+2*3-2+.+n(n+1)-n
=1*2+2*3+...+n(n+1)-(1+2+...+n)
由于n(n+1)=[n(n+1)(n+2)-(n-1)n(n+1)]/3
所以1*2+2*3+...+n(n+1)
=[1*2*3-0+2*3*4-1*2*3+.+n(n+1)(n+2)-(n-1)n(n+1)]/3
[前后消项]
=[n(n+1)(n+2)]/3
所以Sn=1^2+2^2+3^2+.+n^2
=[n(n+1)(n+2)]/3-[n(n+1)]/2
=n(n+1)[(n+2)/3-1/2]
=n(n+1)[(2n+1)/6]
=n(n+1)(2n+1)/6
这个公式在后面常用到
3)1^3+2^3+3^3+.+n^3=( 1+2+3+.+n)^2
=n^2*(n+1)^2÷4
n+1)^4-n^4=[(n+1)^2+n^2][(n+1)^2-n^2]
=(2n^2+2n+1)(2n+1)
=4n^3+6n^2+4n+1
2^4-1^4=4*1^3+6*1^2+4*1+1
3^4-2^4=4*2^3+6*2^2+4*2+1
4^4-3^4=4*3^3+6*3^2+4*3+1
.
(n+1)^4-n^4=4*n^3+6*n^2+4*n+1
各式相加有
(n+1)^4-1=4*(1^3+2^3+3^3...+n^3)+6*(1^2+2^2+...+n^2)+4*(1+2+3+...+n)+n
4*(1^3+2^3+3^3+...+n^3)=(n+1)^4-1+6*[n(n+1)(2n+1)/6]+4*[(1+n)n/2]+n
=[n(n+1)]^2
所以Sn=1^3+2^3+...+n^3=[n(n+1)/2]^2
4)1×2+2×3+3×4+.+n(n+1)
=(1×1+1)+(2×2+2)+(3×3+3)+.(n×n+n)
=(1^2+2^2+3^2+.n^2)+(1+2+3+.n)
=n*(n+1)*(2*n+1)/6+n(n+1)/2
=n(n+1)(n+2)/3
5)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))
=1/4n(n+1)(n+2)(n+3)
这累啊 都没分的 加分再写
1年前
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