(1-1/2²)(1-1/3²)(1-1/4²)……(1-1/2013²)(1-
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(1-1/2²)(1-1/3²)(1-1/4²)……(1-1/2013²)(1-1/2014²)
![(1-1/2²)(1-1/3²)(1-1/4²)……(1-1/2013²)(1-](/uploads/image/z/18256820-68-0.jpg?t=%EF%BC%881-1%2F2%26%23178%3B%EF%BC%89%EF%BC%881-1%2F3%26%23178%3B%EF%BC%89%EF%BC%881-1%2F4%26%23178%3B%EF%BC%89%E2%80%A6%E2%80%A6%EF%BC%881-1%2F2013%26%23178%3B%EF%BC%89%EF%BC%881-)
(1-1/2²)(1-1/3²)(1-1/4²)…(1-1/2012²)(1-2013²)(1-1/2014²)
=(1-1/2)(1+1/2)(1-1/3)(1+1/3)(1-1/4)(1+1/4)×.×(1-1/2013)(1+1/2013)(1-1/2014)(1+1/2014)
=1/2×3/2×2/3×4/3×3/4×5/4×.×2011/2012×2013/2012×2012/2013×2014/2013×2013/2014×2015/2014
=1/2×2015/2014
=2015/4028
=(1-1/2)(1+1/2)(1-1/3)(1+1/3)(1-1/4)(1+1/4)×.×(1-1/2013)(1+1/2013)(1-1/2014)(1+1/2014)
=1/2×3/2×2/3×4/3×3/4×5/4×.×2011/2012×2013/2012×2012/2013×2014/2013×2013/2014×2015/2014
=1/2×2015/2014
=2015/4028
(1-1/2²)(1-1/3²)(1-1/4²)……(1-1/10²)
计算(1-1/2²)(1-1/3²)(1-1/4²)……(1-1/2011²)(
计算:(1-1/2²)(1-1/3²)(1-1/4²)……(1-1/100²)
(1-2²/1)(1-3²/1)(1-4²/1)…(1-9²/1)(1-10
(1)利用因式分解计算:1-2²+3²-4²……+99²-100²+1
(1-2²分之一)(1-3²分之一)(1-4²分之一)……(1-10²分之一
(1-1/2²)(1-1/3²)(1-1/4²)…(1-1/20²)用简便算法计
问道数学题:(1-1/2²)(1-1/3²)(1-1/4²)…(1-1/100²
求和:1²-2²+3²-4²…+(-1)^(n-1)·n²
(1²+3²+…+99²)-(2²+4²+…+100²)=?
(1-1/2²)(1-1/3²)(1-1/3²)……(1-1/10²)
计算:(1²-2²/1+2)+(2²-3²/2+3)+…+(99²-1