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Question Number 109729 by lazygorilla last updated on 25/Aug/20

Answered by prakash jain last updated on 25/Aug/20

1+x+x^2 +..+x^(n−1) =(((x^n −1))/(x−1)) (x≠1) (Geometric Progression) 2^0 +2^1 +...+2^(m−1) =2^m −1 substitute this and you will get the result

1+x+x2+..+xn−1=(xn−1)x−1(x≠1)(GeometricProgression)20+21+...+2m−1=2m−1substitutethisandyouwillgettheresult

Answered by bobhans last updated on 25/Aug/20

consider GP : 1+2+4+8+...+2^(m−1) = ((1(2^m −1))/(2−1)) = 2^m −1

considerGP:1+2+4+8+...+2m−1=1(2m−1)2−1=2m−1

Answered by 1549442205PVT last updated on 25/Aug/20

2^0 +2^1 +2^2 +....+2^(m−1) is the sum of m terms of geometric progress with first term be u_1 =2^0 =1,q=2,so S=u_1 ((q^m −1)/(q−1))=1.((2^m −1)/(2−1))=2^m −1

20+21+22+....+2m−1isthesumofmtermsofgeometricprogresswithfirsttermbeu1=20=1,q=2,soS=u1qm−1q−1=1.2m−12−1=2m−1

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