tag:blogger.com,1999:blog-60761566653515510312015-09-17T02:01:37.541-07:00Math SolutionsI believe there is many ways of solving mathematics. This blog is actually created to share some of my ways of solving of maths.Md Raziunhttp://www.blogger.com/profile/08952187348195074082noreply@blogger.comBlogger1125tag:blogger.com,1999:blog-6076156665351551031.post-26496591747263925132012-01-05T08:51:00.000-08:002012-01-05T08:51:12.739-08:00This was one of the math Olympiad question in 2009<div dir="ltr" style="text-align: left;" trbidi="on">Question<br />(1-1/2^2) (1-1/3^2) (1-1/4^2) ... (1-1/1000^2) = ?<br /><br />Solution<br /><br />The question can be written in this way as well<br />{1- (1/2)^2}{1- (1/3)^2}...<br />= {1+(1/2)} {1-(1/2)}.... and so on ......using (a^2 - b^2) formula<br />= (3/2) (1/2) * (4/3) (2/3) * (5/4) (3/4) .... and so on<br /><br />if we continue this way we see a sequence { (n+1) / n }* { (n-1) / n } = { (n^2 - 1) / n^2)<br /><br />if we find the product of the first four terms we see the result is { (3/4) * (8/9) * (15/16) * (24/25) }= 3/5<br /><br />then the product of the first eight term is 5/9, then the product of the first twelve term is 7/13.. then the product of the first sixteen term is 9/15 and so on....<br /><br />So product of first number of terms, n is [ { (n/2)+1 } / (n + 1) ]<br /><br />So product of first 1000th term is 501/1001 ..<br /><br />Well actually the solution is easy it depends on how fast it clicks on your mind... </div>Md Raziunhttp://www.blogger.com/profile/08952187348195074082noreply@blogger.com0