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下图是6个求和公式.我想知道得出这些求和公式的详细步骤

来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/05/21 10:57:06
下图是6个求和公式.我想知道得出这些求和公式的详细步骤
1.(说明,此题后面打错了.应该是“(ar^(n+1)-a)/(r-1)”)
∵∑ar^k=a+ar+ar²+.+ar^n.(1)
==>r∑ar^k=ar+ar²+ar³+.+ar^n+ar^(n+1).(2)
==>(r-1)∑ar^k=ar^(n+1)-a ((2)式-(1)得)
∴∑ar^k=[ar^(n+1)-a]/(r-1) (∵r≠1);
2.∵1,2,3,.,n是等差数列,而等差数列的求和公式是:等差数列的和=项数×(首项+末项)/2
又等差数列1,2,3,.,n的首项是1,末项是n,项数是n
∴∑k=n(n+1)/2;
3.∵(k+1)³=k³+3k²+3k+1
==>k²=1/3[(k+1)³-k³-3k-1]
∴1²=1/3[2³-1³-3*1-1]
2²=1/3[3³-2³-3*2-1]
3²=1/3[4³-3³-3*3-1]
.
n²=1/3[(n+1)³-n³-3n-1]
故∑k²=1/3[(n+1)³-1-3(1+2+3+.+n)-n] (由前面n个等式相加得)
=1/3[(n+1)³-3n(n+1)/2-(n+1)] (由2题结果得)
=(n+1)/6[2(n+1)²-3n-2] (提取公因式)
=(n+1)(2n²+n)/6
=n(n+1)(2n+1)/6;
4.∵(k+1)^4=k^4+4k³+6k²+4k+1
==>k³=1/4[(k+1)^4-k^4-6k²-4k-1]
∴1³=1/4[2^4-1^4-6*1²-4*1-1]
2³=1/4[3^4-2^4-6*2²-4*2-1]
3³=1/4[4^4-3^4-6*3²-4*3-1]
.
n³=1/4[(n+1)^4-n^4-6n²-4n-1]
故∑k³=1/4[(n+1)^4-1-6(1²+2²+3²+.+n²)-4(1+2+3+.+n)-n] (由前面n个等式相加得)
=1/4[(n+1)^4-n(n+1)(2n+1)-2n(n+1)-(n+1)] (由2题和3题结果得)
=(n+1)/4[(n+1)³-n(2n+1)-2n-1] (提取公因式)
=(n+1)(n³+n²)/4
=n²(n+1)²/4;
5.设Sn=1+x+x²+x³+.+x^n.(1)
∵xSn=x+x²+x³+.+x^n+x^(n+1).(2)
∴由(1)式-(2)式得 (1-x)Sn=1-x^(n+1)
==>Sn=[1-x^(n+1)]/(1-x) (∵│x│∞)Sn
=lim(n->∞){[1-x^(n+1)]/(1-x)}
=(1-0)/(1-x) (∵│x│∞)[x^(n+1)]=0)
=1/(1-x);
6.∵∫[∑kx^(k-1)]dx=∑∫[kx^(k-1)]dx (在收敛区间│x│