Chapter 4 Methods of Induction and Binomial Theorem Ex 4.5 Solutions

Since C0 + C1 + C2 + C3 + ….. + Cn = 2n Putting n = 8, we get C0 + C1 + C2 + ….. + C8 = 28 ∴ C0 + C1 + C2 + ….. + C8 = 256

Since C0 + C1 + C2 + C3 + ….. + Cn = 2n Putting n = 9, we get C0 + C1 + C2 + ….. + C9 = 29 ∴ C0 + C1 + C2 + …… + C9 = 512

Since C0 + C1 + C2 + C3 + ….. + Cn = 2n Putting n = 7, we get C0 + C1 + C2 + ….. + C7 = 27 ∴ C0 + C1 + C2 +….. + C7 = 128 But, C0 = 1 ∴ 1 + C1 + C2 + ….. + C7 = 128 ∴ C1 + C2 + ….. + C7 = 128 – 1 = 127

Since C0 + C1 + C2 + C3 + ….. + Cn = 2n Putting n = 6, we get C0 + C1 + C2 + ….. + C6 = 26 ∴ C0 + C1 + C2 + …… + C6 = 64 But, C0 = 1 ∴ 1 + C1 + C2 + ….. + C6 = 64 ∴ C1 + C2 + ….. + C6 = 64 – 1 = 63

Since C0 + C1 + C2 + C3 + …… + Cn = 2n Putting n = 8, we get C0 + C1 + C2 + C3 + …… + C8 = 28 But, sum of even coefficients = sum of odd coefficients ∴ C0 + C2 + C4 + C6 + C8 = C1 + C3 + C5 + C7 Let C0 + C2 + C4 + C6 + C8 = C1 + C3 + C5 + C7 = k Now, C0 + C1 + C2 + C3 + C4 + C5 + C6 + C7 + C8 = 256 ∴ (C0 + C2 + C4 + C6 + C8) + (C1 + C3 + C5 + C7) = 256 ∴ k + k = 256 ∴ 2k = 256 ∴ k = 128 ∴ C0 + C2 + C4 + C6 + C8 = C1 + C3 + C5 + C7 = 128

Since C0 + C1 + C2 + C3 + ….. + Cn = 2n But, C0 = 1 ∴ 1 + C1 + C2 + C3 + …… + Cn = 2n ∴ C1 + C2 + C3 + ….. + Cn = 2n – 1