# NCERT Solutions For Class 11 Maths Binomial Theorem ## NCERT Solutions For Class 11 Maths Binomial Theorem

Maths NCERT Solutions Class 11 : Suppose you need to calculate the amount of interest you will get after 5 years on a sum

of money that you have invested at the rate of 15% compound interest per year. Or

suppose we need to find the size of the population of a country after 10 years if we know

the annual growth rate. A result that will help in finding these quantities is the binomial

theorem. This theorem, as you will see, helps us to calculate the rational powers of any

real binomial expression, that is, any expression involving two terms.

n your daily interactions, you must have made several assertions in the form of sentences. Of

these assertions, the ones that are either true or false are called statement or propositions.

For instance,

“I am 20 years old” and “If x = 3, then x2 = 9” are statements, but ‘When will you leave?’ And

‘How wonderful!’ are not statements.

Notice that a statement has to be a definite assertion which can be true or false, but not both.

For example, ‘x – 5 = 7’ is not a statement because we don’t know what x, is. If x = 12, it is

true, but if x = 5, ‘it is not true. Therefore, ‘x – 5 = 7’ is not accepted by mathematicians as a

statement.

But both ‘x – 5 = 7 ⇒x = 12’ and x – 5 = 7 for any real number x’ are statements, the first one

true and the second one false.

Maths NCERT Solutions Class 11 are available, click CBSE Class 11th Maths for details.

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Ques. If P (n) denotes 2n> n–1, write P (1), P (k) and P (k+1), where k ∈ N.

Solution: Replacing n by 1, k and k + 1, respectively in P (n),

We get P (1) : 21> 2 – 1, i.e., 2 > 1

P (k) : 2k> k – 1

P (k + 1) : 2k+1> (k + 1) – 1, i.e., 2k+1> k

Ques. If P (n) is the statement

‘1 + 4 + 7 + (3n – 2) = n(3n − 1)/2

Write P (1), P(k) and P(k + 1).

Solution: To write P(1), the terms on the left-hand side (LHS) of P(n) continue till

3× 1 – 2, i.e., 1. So, P (1) will have only one term in its LHS, i.e., the first term.

Also, the right hand side (RHS) of P(1) = 1*(3*1-1) / 2

= 1

Therefore, P (1) is 1 = 1.

Replacing n by 2, we get

P(2) : 1 + 4 = 2*(3*2-1) / 2

i.e., 5 = 5.

Replacing n by k and k + 1, respectively, we get

P(k) : 1 + 4 + 7 + …. + (3k – 2)

=  k(3k – 1) / 2

P(k + 1) : 1 + 4 + 7 + …. + (3k – 2) + [3 (k + 1) – 2]

=(k+1)[3(k+1)-1] / 2

i.e. , 1 + 4 + 7 +…. + (3k + 1) =(k + 1)[(3k + 2) / 2.

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After studying this lesson, you will be able to:

• state the Principle of (finite) Mathematical Induction;
• verify the truth or otherwise of the statement P(n) for n= 1;
• verify if P(k+1) is true, assuming that P(k) is true;
• use the principle of mathematical induction to establish the truth or otherwise of mathematical statements;
• state the binomial theorem for a positive integral index and prove it using the principle of mathematical induction;
• write the binomial expansion for expressions like ( x + y ) n for different values of x and y using binomial theorem;
• write the general term and middle term (s) of a binomial expansion;
• write the binomial expansion for negative as well as for rational indices;
• apply the binomial expansion for finding approximate values of numbers like 3√9, √2,  3√3 etc; and
• apply the binomial expansion to evaluate algebraic expressions like ( 3 – 5/x)7, where x is so small that 2 x and higher powers of x can be neglected.

To be continued…..

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August 26, 2019

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