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 C_{0} + C_{1} + C_{2} + ….. + C_{8} = 256

Solution:

Since C_{0} + C_{1} + C_{2} + C_{3} + ….. + C_{n} = 2^{n}

Putting n = 8, we get

C_{0} + C_{1} + C_{2} + ….. + C_{8} = 2^{8}

∴ C_{0} + C_{1} + C_{2} + ….. + C_{8} = 256

Question 2.

Show that C_{0} + C_{1} + C_{2} + …… + C_{9} = 512

Solution:

Since C_{0} + C_{1} + C_{2} + C_{3} + ….. + C_{n} = 2^{n}

Putting n = 9, we get

C_{0} + C_{1} + C_{2} + ….. + C_{9} = 2^{9}

∴ C_{0} + C_{1} + C_{2} + …… + C_{9} = 512

Question 3.

Show that C_{1} + C_{2} + C_{3} + ….. + C_{7} = 127

Solution:

Since C_{0} + C_{1} + C_{2} + C_{3} + ….. + C_{n} = 2^{n}

Putting n = 7, we get

C_{0} + C_{1} + C_{2} + ….. + C_{7} = 2^{7}

∴ C_{0} + C_{1} + C_{2} +….. + C_{7} = 128

But, C_{0} = 1

∴ 1 + C_{1} + C_{2} + ….. + C_{7} = 128

∴ C_{1} + C_{2} + ….. + C_{7} = 128 – 1 = 127

Question 4.

Show that C_{1} + C_{2} + C_{3} + ….. + C_{6} = 63

Solution:

Since C_{0} + C_{1} + C_{2} + C_{3} + ….. + C_{n} = 2^{n}

Putting n = 6, we get

C_{0} + C_{1} + C_{2} + ….. + C_{6} = 2^{6}

∴ C_{0} + C_{1} + C_{2} + …… + C_{6} = 64

But, C_{0} = 1

∴ 1 + C_{1} + C_{2} + ….. + C_{6} = 64

∴ C_{1} + C_{2} + ….. + C_{6} = 64 – 1 = 63

Question 5.

Show that C_{0} + C_{2} + C_{4} + C_{6} + C_{8} = C_{1} + C_{3} + C_{5} + C_{7} = 128

Solution:

Since C_{0} + C_{1} + C_{2} + C_{3} + …… + C_{n} = 2^{n}

Putting n = 8, we get

C_{0} + C_{1} + C_{2} + C_{3} + …… + C_{8} = 2^{8}

But, sum of even coefficients = sum of odd coefficients

∴ C_{0} + C_{2} + C_{4} + C_{6} + C_{8} = C_{1} + C_{3} + C_{5} + C_{7}

Let C_{0} + C_{2} + C_{4} + C_{6} + C_{8} = C_{1} + C_{3} + C_{5} + C_{7} = k

Now, C_{0} + C_{1} + C_{2} + C_{3} + C_{4} + C_{5} + C_{6} + C_{7} + C_{8} = 256

∴ (C_{0} + C_{2} + C_{4} + C_{6} + C_{8}) + (C_{1} + C_{3} + C_{5} + C_{7}) = 256

∴ k + k = 256

∴ 2k = 256

∴ k = 128

∴ C_{0} + C_{2} + C_{4} + C_{6} + C_{8} = C_{1} + C_{3} + C_{5} + C_{7} = 128

Question 6.

Show that C_{1} + C_{2} + C_{3} + ….. + C_{n} = 2^{n} – 1

Solution:

Since C_{0} + C_{1} + C_{2} + C_{3} + ….. + C_{n} = 2^{n}

But, C_{0} = 1

∴ 1 + C_{1} + C_{2} + C_{3} + …… + C_{n} = 2^{n}

∴ C_{1} + C_{2} + C_{3} + ….. + C_{n} = 2^{n} – 1

Question 7.

Show that C_{0} + 2C_{1} + 3C_{2} + 4C_{3} + ….. + (n + 1)C_{n} = (n + 2) 2^{n-1}

Solution: