Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

$\sum_{n=2}^{100}\; \log \frac{n-1}{n}$ is gelijk aan	A. − 1
B. 2
C. − 2
D. 100
E. log 100!

$\sum_{n=2}^{100}\; \log \frac{n-1}{n}$ is equal to	A. − 1
B. 2
C. − 2
D. 100
E. log 100!

1^ste manier :
$\sum_{n=2}^{100}\;\log\frac{n-1}{n}=\log\frac12+\log\frac23+\log\frac34+\cdots\log\frac{98}{99}+\log\frac{99}{100}\\=(\log1-\log2)+(\log2-\log3)+(\log3-\log4)+\cdots+(\log 98-\log99)\\+(\log 99-\log 100)=\log1-\log100=-\log 100=-2$
2^de manier :
$\sum_{n=2}^{100}\;\log\frac{n-1}{n}=\log\frac12+\log\frac23+\log\frac34+\cdots\log\frac{98}{99}+\log\frac{99}{100}\\=\log\left(\frac12.\frac23.\frac34.\cdots.\frac{98}{99}.\frac{99}{100}\right)=\log\frac{1}{100}=\log1-\log100=-\log100=-2$