Grijesh Chauhan
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 Feb 14 comment What's the intuition behind Pythagoras' theorem? @j08691 I like your image and voting your comment as an answer. Thanks! Feb 14 comment What's the intuition behind Pythagoras' theorem? I was about to ask this question. :) Jan 13 comment Calculating the number of possible paths through some squares interesting thanks! Jul 29 comment Binary to Decimal Thanks Gammatester! Jul 29 comment Binary to Decimal Thanks for the answer! but I want to wait for few days before accept an answer. Jul 29 comment Binary to Decimal @RGB Yes If I am given 8 then I need 1000, but additionally I wants to convert without loop. (and a power function need a loop). Can we have some mathematical equation/function such that if f(8) == 1000 some liner equation? .... yet I think its not possible. Jul 29 comment Binary to Decimal @gammatester Can you post an answer with a more description, I will be so grateful. Jul 29 comment Binary to Decimal @RGB As I respond to Adrino, I am only interested in patterns that starts with 1 followed by 0s for example 1, 10, 100, 1000, I don't need 101 as input. -- What I wants if I have binary 100 that is actually 4 then I need 100 hundred, similarly if I have 1000 that is actually 8 then I need 1000 thousand. --- After some comments and try I feel that what I need is not possible. Jul 29 comment Binary to Decimal @BenMillwood its more mathematical question I think. I want an equation for this restrict problem domain. --- Jul 29 comment Binary to Decimal @Adriano It will need a loop, I am looking for equation instead. Otherwise power() function I could have use. Yes I need equation for this particular pattern, I am not interested in binary 1010 -- decimal 1010 Jul 29 comment Binary to Decimal I need it in computer-programming Jun 10 comment How to simplify summation $(1\cdot2) + (2\cdot3) + (3\cdot4) + (4\cdot5) + (5\cdot6) + … + (N\cdot(N+1))$ in terms of N? @ByronSchmuland oh! this because I partially copied from linked question, Now corrected. Thanks :) Jun 10 comment How to simplify summation $(1\cdot2) + (2\cdot3) + (3\cdot4) + (4\cdot5) + (5\cdot6) + … + (N\cdot(N+1))$ in terms of N? ok So you updated your answer.. its correct na Thanks a lot vonbrand! Jun 10 comment How to simplify summation $(1\cdot2) + (2\cdot3) + (3\cdot4) + (4\cdot5) + (5\cdot6) + … + (N\cdot(N+1))$ in terms of N? @MarkBennet Got it!..done thanks :) Jun 10 comment How to simplify summation $(1\cdot2) + (2\cdot3) + (3\cdot4) + (4\cdot5) + (5\cdot6) + … + (N\cdot(N+1))$ in terms of N? @MarkBennet Please check my answer Jun 10 comment How to simplify summation $(1\cdot2) + (2\cdot3) + (3\cdot4) + (4\cdot5) + (5\cdot6) + … + (N\cdot(N+1))$ in terms of N? Please check my answer .. Jun 10 comment How to simplify summation $(1\cdot2) + (2\cdot3) + (3\cdot4) + (4\cdot5) + (5\cdot6) + … + (N\cdot(N+1))$ in terms of N? Hi please read Edit part its actually: (1*2) + (3*2) + (4*3) + (5*4) + (6*5) + ... + (N * (N+1)) Sorry I mistake I made Jun 10 comment How to simplify summation $(1\cdot2) + (2\cdot3) + (3\cdot4) + (4\cdot5) + (5\cdot6) + … + (N\cdot(N+1))$ in terms of N? @MarkBennet Got it Thanks!!, But I am sure for my final sequence (1*2) + (3*2) + (4*3) + (5*4) + (6*5) + ... + (N * (N-1)) Jun 10 comment How to simplify summation $(1\cdot2) + (2\cdot3) + (3\cdot4) + (4\cdot5) + (5\cdot6) + … + (N\cdot(N+1))$ in terms of N? do you means (1*2) + (3*2) + (4*3) + (5*4) + (6*5) + ... + (N * (N-1)) ? by last description Jun 10 comment How to simplify summation $(1\cdot2) + (2\cdot3) + (3\cdot4) + (4\cdot5) + (5\cdot6) + … + (N\cdot(N+1))$ in terms of N? @MarkBennet 6 is absent in series