2 votes 2 votes please share some resource also; so that i can see the similar identities! Gate Fever asked Oct 31, 2018 Gate Fever 507 views answer comment Share Follow See 1 comment See all 1 1 comment reply kumar.dilip commented Oct 31, 2018 reply Follow Share n.2n-1 https://www.algebra.com/algebra/homework/equations/Equations.faq.question.1093265.html 1 votes 1 votes Please log in or register to add a comment.
0 votes 0 votes Hope it helps. manohar answered Oct 31, 2018 manohar comment Share Follow See all 5 Comments See all 5 5 Comments reply Gate Fever commented Oct 31, 2018 reply Follow Share your explanation is still not clear to me!! plus the image is also blurred!! 0 votes 0 votes manohar commented Oct 31, 2018 reply Follow Share Hey sorry about that. Moral of the story is that consecutive terms form a simple pattern. First convert r.nCr to (n-r+1).nCr-1. Then add both series, one in nCr form and other in nCr-1 form. If you try yourself, you will get my answer. Math is not a subject to read, try yourself and ask where you are stuck. Sorry for the pic. 0 votes 0 votes Gate Fever commented Oct 31, 2018 reply Follow Share r.nCr to (n-r+1).nCr-1. how u thought this?? i mean is it a identity, or u created while just solving the ques?? because it would take me months to think like this and put value 0 votes 0 votes manohar commented Oct 31, 2018 reply Follow Share r.n!/r!.(n-r)! =n!/(r-1)!.(n-r)! =n!.(n-r+1)/(r-1)!.[n-(r-1)]! =(n-r+1).nCr-1. You don't have to memorize it, but with enough practice you will find solutions easily. 0 votes 0 votes manohar commented Oct 31, 2018 reply Follow Share You can practice summations in 12th class books meant for JEE. (Like Cengage Algebra). Here is an article for you https://www.askiitians.com/iit-jee-progressions-and-series/method-of-differences/ 0 votes 0 votes Please log in or register to add a comment.
