作业帮求解GMAT一道数学题[1/(2n-1)!]-[1/(2n+1)!]简化后等于an^2+bn+c,问a+b+c等于多少?
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求解GMAT一道数学题[1/(2n-1)!]-[1/(2n+1)!]简化后等于an^2+bn+c,问a+b+c等于多少?
求解GMAT一道数学题
[1/(2n-1)!]-[1/(2n+1)!]简化后等于an^2+bn+c,问a+b+c等于多少?
求解GMAT一道数学题[1/(2n-1)!]-[1/(2n+1)!]简化后等于an^2+bn+c,问a+b+c等于多少?
这道题其实没那么复杂,不需要进行化简.
只要令n=1,则an^2+bn+c=a+b+c
那么当n=1时,原式=[1/(2-1)!]-[1/(2+1)!]=1-1/6=5/6
所以a+b+c=5/6
PS:我一开始也抓破头皮想化简,可既然题目是说化简后一定为该式,那么不用管它是如何化简的,因为两式是等价的,再观察后两式,便可发现上述解法
原式=[2n(2n+1)-1]/(2n+1)!
=(4n^2+2n-1)/(2n+1)!
=[4/(2n+1)!]n^2+[2/(2n+1)!]n-1/(2n+1)!
所以a=4/(2n+1)!,b=2/(2n+1)!,c=-1/(2n+1)!
所以a+b+c=5/(2n+1)!
这道题目应该是
[1/(2n-1)!]-[1/(2n+1)!]简化后等于an^2+bn+c/[1/(2n+1)!]，问a+b+c等于多少？
这样的话就可以得出a+b+c=5
[1/(2n-1)!]-[1/(2n+1)!]=4n^2+2n-1/[1/(2n+1)!] a=4,b=2,c=-1 所以a+b+c=5
求解GMAT一道数学题[1/(2n-1)!]-[1/(2n+1)!]简化后等于an^2+bn+c,问a+b+c等于多少?
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