下面级数求和错误的是 A: $\sum_{n=0}^\infty q^n = \frac{1}{1-q} (0\lt q\lt1) $ B: $\sum_{n=1}^\infty \frac{x^{2^{n-1}}}{1-x^{2^n}} = \frac{x}{1-x} (|x|\lt 1) $ C: $\sum_{n=1}^\infty \frac{1}{{n!}} = e $ D: $\sum_{n=1}^\infty \frac{x^{2^{n-1}}}{1-x^{2^n}} = \frac{1}{1-x} (x>1) $

Solve $\sum_{n=1}^{\infty}\frac{2n-1}{2^n}$:<br/>______

Solve $\sum_{n=1}^{\infty}\frac{1}{n(n+1)(n+2)}$:<br/>______

设$\{a_n\}$是正项数列，则下列选项中正确的是 A: 若$a_n&gt;a_{n+1}$，则$\sum_{n=1}^{\infty}(-1)^{n-1}a_n$收敛 B: 若$\sum_{n=1}^{\infty}(-1)^{n-1}a_n$收敛，则$a_n&gt;a_{n+1}$ C: 若$\sum_{n=1}^{\infty}a_n$收敛，则存在常数$p&gt;1$，使得$\lim_{n\to\infty}n^pa_n$存在 D: 若存在常数$p&gt;1$，使得$\lim_{n\to\infty}n^pa_n$存在，则$\sum_{n=1}^{\infty}a_n$收敛

函数$f(x)=\arcsin(\sin x)$的傅里叶级数展开式为 A: $x$ B: $$\frac{4}{\pi}\sum_{n=0}^{\infty}\frac{(-1)^n\sin(2n+1)x}{(2n+1)^2}$$ C: $$\frac{4}{\pi}\sum_{n=1}^{\infty}\frac{(-1)^n\sin(2n+1)x}{(2n+1)^2}$$ D: $$\frac{4}{\pi}\sum_{n=1}^{\infty}\frac{(-1)^{n-1}\sin(2n+1)x}{(2n+1)^2}$$

def func(): global sum sum=0 print(sum) for i in range(5): sum+=1 print(sum) func() print(sum)

阅读下面代码，判断正误sum = 0i = 1while i<;10: sum = sum + i i = i +1print "sum is ", sum

将下面的for循环转换为while循环，补全代码。 long sum=0; for(inti=0;i<=100;i++) sum=sum+i; System.out.println("sum="+sum); 改写成： long sum=0; ① ； while( ② ){ sum=sum+i; ③ ; } System.out.println("sum="+sum);

System.out.println("sum:"+sum);为输出sum的值。

在下列程序段中，不能计算1到100之间奇数之和的是__________。 A: Dim sum As Integer sum = 0 For i = 1 To 100 Step 2 sum = sum + i Next B: Dim sum As Integer sum = 0 For i = 1 To 100 If i Mod 2 &lt;&gt; 0 Then sum = sum + i Next C: Dim sum As Integer sum = 0 For i = 1 To 99 sum = sum + i Next D: Dim sum As Integer sum = 0 For i = 100 To 1 Step -1 If i Mod 2 &lt;&gt; 0 Then sum = sum + i Next