Question

Cho P=\(\left(\frac{1}{2^2}-1\right).\left(\frac{1}{3^2}-1\right).\left(\frac{1}{4^2}-1\right)...\left(\frac{1}{100^2}-1\right).\)

So sánh P với \(\frac{-1}{2}\)

Answer 1

Lời giải:

Ta có:

\(P=\frac{1-2^2}{2^2}.\frac{1-3^2}{3^2}.\frac{1-4^2}{4^2}...\frac{1-100^2}{100^2}=-\frac{(2^2-1)(3^2-1)(4^2-1)...(100^2-1)}{2^2.3^2.4^2....100^2}\)

\(=-\frac{(2-1)(2+1)(3-1)(3+1)(4-1)(4+1)....(100-1)(100+1)}{(2.3.4..100)(2.3.4...100)}\)

\(=-\frac{[(2-1)(3-1)....(100-1)][(2+1)(3+1)....(100+1)]}{(2.3....100)(2.3.4....100)}\)

\(=-\frac{(1.2.3...99)(3.4.5...101)}{(2.3....100)(2.3....100)}=-\frac{1.2.3...99}{2.3.4..100}.\frac{3.4.5....101}{2.3.4..100}\)

\(=-\frac{1}{100}.\frac{101}{2}=\frac{-101}{200}<\frac{-100}{200}\)

Hay $P<\frac{-1}{2}$

Answer 2

Lời giải:

Ta có:

\(P=\frac{1-2^2}{2^2}.\frac{1-3^2}{3^2}.\frac{1-4^2}{4^2}...\frac{1-100^2}{100^2}=-\frac{(2^2-1)(3^2-1)(4^2-1)...(100^2-1)}{2^2.3^2.4^2....100^2}\)

\(=-\frac{(2-1)(2+1)(3-1)(3+1)(4-1)(4+1)....(100-1)(100+1)}{(2.3.4..100)(2.3.4...100)}\)

\(=-\frac{[(2-1)(3-1)....(100-1)][(2+1)(3+1)....(100+1)]}{(2.3....100)(2.3.4....100)}\)

\(=-\frac{(1.2.3...99)(3.4.5...101)}{(2.3....100)(2.3....100)}=-\frac{1.2.3...99}{2.3.4..100}.\frac{3.4.5....101}{2.3.4..100}\)

\(=-\frac{1}{100}.\frac{101}{2}=\frac{-101}{200}<\frac{-100}{200}\)

Hay $P<\frac{-1}{2}$