## Rajasthan Board RBSE Class 11 Maths Chapter 7 Binomial Theorem Ex 7.4

Question 1.

Expand the following Binomials upto fourth term:

Solution:

Expansion upto four terms of (1 + x^{2})^{-2
}Question 2.

Find the required terms in the following expansions:

(i) Fourth term of (1 – 3x)^{-1/3
}(ii) Seventh term of (1 + x)^{5/2
}(iii) Eighth term of (1 + 2x)^{-1/2
}Solution:

(i) Fourth term of (1 – 3x)^{-1/3
}= Fourth term of (1 – 3x)^{-1/3}^{
}= Fourth term of {1 + (- 3x)}^{– }^{1/3}

(ii) Seventh term of (1 + x)^{5/2
}

(iii) Eighth term of (1 + 2x)^{-1/2
}

Question 3.

Find the general term of the following expansions:

(i) (a^{3} – x^{3})^{2/3}

(ii) (1 – 2x)^{-3/2}^{
}(iii) (1 – x)^{-p/q
}Solution:

(i) General term in the expansion of (a^{3} – x^{3})^{2/3
}

(ii) General term in the expansion of (1 – 2x)^{-3/2
}

(iii) General term of in the expansion of (1 – x)^{-p/q}

Question 4.

If x < 3, find the coefficient of (3 – x)^{-8} in the expansion of (3 – x)^{– 5}.

Solution:

(x + 1)th term in the expansion of (3 – x)^{-8
}

Question 5.

Find the coefficient of x^{6} in the expansion of (a + 2bx^{2})^{-3
}Solution :

(r + 1)th term in the expansion of (a + 2bx^{2})^{-3
}

For the coefficient of x^{6} in the this term

2r = 6 ⇒ r = 3

Hence, from equation (i), coefficient of x^{6} in the expansion of (a + 2bx^{2})^{-3
}

Question 6.

Find the coefficient of x^{10} in the expansion of

Solution:

Question 7.

Find the coefficient of x^{r} in the expansion of (1 – 2x + 3x^{2} – 4x^{3} + …)^{n} and if x = \(\frac { 1 }{ 2 }\) and n = 1, then find the value of the expression.

Solution:

we know that (1 + x)^{-2} = 1 – 2x + 3x^{2} – 4x^{3} +…+ (-1)^{r} (r+1)x^{r}+ ….. (i)

Given progression (1 – 2x + 3x^{2} – 4x^{3} +…)^{n}

(1 – 2x + 3x^{2} – 4x^{3} +…)^{n
}= {(1 + x)^{-2}}^{n }[From equation (i)]

= (1 + x)^{-2n
}

Question 8.

Prove that (1 + x + x^{2} + x^{3} + …)^{2} = 1 + 2x + 3x^{2} + …

Solution:

We know that

1 + x + x^{2} + x^{3} + … = (1 – x)^{-1} …. (i)

(1 + x + x^{2} + x^{3} + …)^{2
}= {(1 – x)^{-1}}^{2 }[From equation (i)]

= (1 – x)^{2}

Question 9.

Prove that (1 + x + x^{2} + x^{3} +….) (1 + 3x + 6x^{2}+ …)

= (1 + 2x + 3x^{2} + …)^{2
}Solution:

We know that

(1 – x)^{-1} = 1 + x + x^{2} + x^{3} +… … (i)

and (1 – x)^{-3} = 1 + 3x + 6x^{2} + 10x^{3} + … (ii)

L.H.S. = (1 + x + x^{2} + x^{3} +…)(1 + 3x + 6x^{2} + …)

= {(1 – x)^{-1}} {(1 – x)^{-3}}

= {(1 – x)^{2}}^{-2
}= {(1 – x)^{-2}}^{2
}= (1 + 2x + 3x^{2} +…)^{2
}= R.H.S.

Hence Proved.

Question 10.

If x = 2y + 3y^{2} + 4y^{3} + …then express y in series of as ascending powers of x.

Solution:

⇒ x = 2y + 3y^{2} + 4y^{2} + …

⇒ x = (1 + 2y + 3y^{2} + 4y3 + …)- 1

[∵ (1 – x)^{-2 }= 1 + 2x + 3x^{2} + …]

⇒ (x+ 1) = (1 -y)^{-2
}

## Leave a Reply