Ques. from bitsat sample paper...?

Solve :

²ⁿ⁺¹C₁ + ²ⁿ⁺¹C₂ + ²ⁿ⁺¹C₃ + ··· + ²ⁿ⁺¹Cn = 1023.

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There may be a better way to do this, but here is one approach. Look for a pattern with small n values and then infer from that.

n = 1 ----> 3 books. Can choose up to 1 book in 3C1 = 3 ways.

n = 2 ----> 5 books. Can choose up to 2 books in 5C1 + 5C2 = 15 ways

n = 3 ----> 7 books. Can choose up to 3 books in 7C1 + 7C2 + 7C3 = 63 ways.

3 = 2^2 - 1

15 = 2^4 - 1

63 = 2^6 - 1

The pattern seems to be 2^(2n) - 1

Can you finish from there?

2n+1c1=1023

2n+1=1023

2n=1022

n=511

(2n+1)! /n!*n+1! = 1023