{eq}\sum_{i=1}^{n}i^2=\dfrac{1}{6}n(n+1)(2n+1) {/eq} Then, solve the obtained limit expression (in terms of the variable n) using the appropriate division rule. {eq}\displaystyle \frac{an^2+bn+c}{n^a}=\frac{an^2}{n^a}+\frac{bn}{n^a}+\frac{c}{...
{1}b_{1}+a_{1}b_{2}+ \cdots +a_{n}b_{n}=|A|B| \cos \thet a S| A \parallel B $$ g(x)$$ \frac{n}{2+} a_{2}b_{2}\sqrt{5}a_{2}^{2}\cdot \sqrt{5} b $$不个等时,可出学归纳值得$$ AA=(1,1) $$幂级数,两平方即可$$ 之 $$杨西 向特...
{ 1 } ( v $$ 感觉应该选C$$ , F $$但Rn(x)趋于0C$$ \lim _ { x \rightarrow \infty } \frac { R _ { x } ( x ) } { x _ { n } y ^ { n - 1 } } = 0 $$ D$$ \lim _ { x \rightarrow \infty } \frac { R _ { n } ( x ) } { ( x - x...
∑n=0∞(1+(−1)n)converges. True False Explain. n′ ∑n=1∞an {an} limn→∞an=0. {bn} ∑n=1∞bn Answer and Explanation:1 The given series is∑n=0∞(1+(−1)n). Let,an=(1+(−1)n). Then,a2n=2and... Learn more about this topic: ...
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Answer to: Find the sum \sum_{k=1}^\infty \frac{k}{(k^2 +1)^3} By signing up, you'll get thousands of step-by-step solutions to your homework...
$$ \lim _ { n \rightarrow \infty } \sqrt [ n ] { u _ { n } } = \lim _ { n \rightarrow \infty } \sqrt [ n ] { ( \frac { n } { 2 n + 1 } ) ^ { n } } = \lim _ { n \rightarrow \infty } \frac { n } { 2 n + 1 } = \frac { 1 ...
$$ 故有 $$ \phi _ { n } ( t ) = \frac { 1 } { n ! } \sum _ { k = 0 } ^ { h } \begin{pmatrix} n \\ 2 k \end{pmatrix} B _ { 2 k } ^ { n } t ^ { n - 2 k } , $$ 其中λ是$$ \frac { 1 } { 2 } ( n + 1 ) $$的整数...
思路 和玩游戏一题类似 定义\(A_k(x)=\sum_{i=0}^\infty a_k^ix^i=\frac{1}{1-a_kx}\) 用\(\ln 'x\)代替\(\frac{1}{x}\), 所以就是求 \[ f(x)=\sum_{i=1}^n \ln'(1-a_ix) \] 这样没法快速计算 所以再设\(G(x)=\sum _{i=1}^n (ln(1-a_ix))'\) 所以 \[ ...
解答解:(Ⅰ)设等比数列{an}的公比为q,由题意可知2a4=2a5+4a6,即a4=a4q+2a4q2, 由an>0,则2q2+q-1=0,解得:q=1212,或q=-1(舍去), a4=4a32=4a2a4,则a2=1414, ∴a1=1212, 等比数列{an}通项公式an=(1212)n, 当n≥2时,bn=Sn-Sn-1=(n+1)bn2(n+1)bn2-nbn−12nbn−12, ...