计算:12-22+32-42+52-62+…+n2-(n+1)2=________.(n属于奇数)
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计算:12-22+32-42+52-62+…+n2-(n+1)2=________.(n属于奇数)
计算:12-22+32-42+52-62+…+n2-(n+1)2=________。(n属于奇数) 每个数字后面的2是平方的意思
计算:12-22+32-42+52-62+…+n2-(n+1)2=________。(n属于奇数) 每个数字后面的2是平方的意思
1²-2²+3²-4²+……n²-(n-1)²
=(1-2)(1+2)+(3-4)(3+4)+……+[n-(n+1)][n+(n+1)]
=(-3)+(-7)+……+[-(2n+1)]
=-3+[(2n+2)/4-1]×(-4)
=-3+[-(2n+2)+4]
=-3+(-2n+2)
=-2n-1
再问: 错了,n=3时,1²-2²+3²-4²=-10 -2n-1=-7
再答: 1²-2²+3²-4²+……n²-(n-1)²
=(1-2)(1+2)+(3-4)(3+4)+……+[n-(n+1)][n+(n+1)]
=(-3)+(-7)+……+[-(2n+1)]
=[(2n+2)/4]×(-3)+{[(2n+2)/4][(2n+2)/4-1]×(-4)}/2
=[-3(n+1)/2]-[(n+1)/2](n-1)
=[-(n+1)/2][3+(n-1)]
=[-(n+1)/2](n+2)
=-(n+1)(n+2)/2
=(1-2)(1+2)+(3-4)(3+4)+……+[n-(n+1)][n+(n+1)]
=(-3)+(-7)+……+[-(2n+1)]
=-3+[(2n+2)/4-1]×(-4)
=-3+[-(2n+2)+4]
=-3+(-2n+2)
=-2n-1
再问: 错了,n=3时,1²-2²+3²-4²=-10 -2n-1=-7
再答: 1²-2²+3²-4²+……n²-(n-1)²
=(1-2)(1+2)+(3-4)(3+4)+……+[n-(n+1)][n+(n+1)]
=(-3)+(-7)+……+[-(2n+1)]
=[(2n+2)/4]×(-3)+{[(2n+2)/4][(2n+2)/4-1]×(-4)}/2
=[-3(n+1)/2]-[(n+1)/2](n-1)
=[-(n+1)/2][3+(n-1)]
=[-(n+1)/2](n+2)
=-(n+1)(n+2)/2
计算:12-22+32-42+52-62+…+n2-(n+1)2=________.(n属于奇数)
12+22+32+42+52+62+72+82+…+n2=n(n+1)(2n+1)/6
12+22+32+42+……+n2=n+(n+1)(2n+1)/6为什么?
证明:12+22+32+……+n2=1/6n(n+1)(2n+1)
已知12+22+32+42+52+-------+n2=6分之1XnX(n+1)(2n+1)试求22+42+62+---
已知函数f(n)=n2,当n为奇数时−n2,当n为偶数时且an=f(n)+f(n+1),则a1+a2+a3+…+a100
若f(n)=√n2+1 -n ,g(n)=n-√n2-1,c(n)=1/2n(n大于或等于2且属于正实数)
请问如何证明lim(n→∞)[n/(n2+n)+n/(n2+2n)+…+n/(n2+nn)]=1,
设f(1)=2,f(n)>0(n属于正整数)有f(n1+n2)=f(n1)f(n2),求f(n)
已知12+22+32+….n2=(1/6)n(n+1)(2n+1),试求22+42+62….+502的值(写过程)
已知:12+22+32+…+n2=1/6n(n+1)(2n+1),试求22+42+62+…+1002
大一求极限lim(n/(n2+1)+n/(n2+2^2)+……+n/(n2+n2))