Lời giải:
Ta có:
\(\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)...........\left(\frac{1}{50^2}-1\right)\)
\(=\frac{(1-2^2)(1-3^2)(1-4^2)...(1-50^2)}{2^2.3^2....50^2}\)
\(=-\frac{(2^2-1)(3^2-1)(4^2-1)...(50^2-1)}{2^2.3^2...50^2}\)
\(=-\frac{(2-1)(2+1)(3-1)(3+1)(4-1)(4+1)...(50-1)(50+1)}{(2.3.4...50)(2.3.4...50)}\)
\(=-\frac{(2-1)(3-1)....(50-1)}{2.3.4...50}.\frac{(2+1)(3+1)....(50+1)}{2.3.4...50}\)
\(=-\frac{1.2.3...49}{2.3.4...50}.\frac{3.4.5..51}{2.3.4..50}=-\frac{1}{50}.\frac{51}{2}=-\frac{51}{100}\)
Bạn chú ý lần sau gõ đề bằng công thức toán!
Đúng 0
Bình luận (0)
