# Maharashtra Board 11th Maths Solutions Chapter 4 Methods of Induction and Binomial Theorem Ex 4.5

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 4 Methods of Induction and Binomial Theorem Ex 4.5 Questions and Answers.

## Maharashtra State Board 11th Maths Solutions Chapter 4 Methods of Induction and Binomial Theorem Ex 4.5

Question 1.
Show that C0 + C1 + C2 + ….. + C8 = 256
Solution:
Since C0 + C1 + C2 + C3 + ….. + Cn = 2n
Putting n = 8, we get
C0 + C1 + C2 + ….. + C8 = 28
∴ C0 + C1 + C2 + ….. + C8 = 256

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

Question 3.
Show that C1 + C2 + C3 + ….. + C7 = 127
Solution:
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

Question 4.
Show that C1 + C2 + C3 + ….. + C6 = 63
Solution:
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

Question 5.
Show that C0 + C2 + C4 + C6 + C8 = C1 + C3 + C5 + C7 = 128
Solution:
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

Question 6.
Show that C1 + C2 + C3 + ….. + Cn = 2n – 1
Solution:
Since C0 + C1 + C2 + C3 + ….. + Cn = 2n
But, C0 = 1
∴ 1 + C1 + C2 + C3 + …… + Cn = 2n
∴ C1 + C2 + C3 + ….. + Cn = 2n – 1

Question 7.
Show that C0 + 2C1 + 3C2 + 4C3 + ….. + (n + 1)Cn = (n + 2) 2n-1
Solution: