## Rajasthan Board RBSE Class 11 Maths Chapter 7 Binomial Theorem Ex 7.3

Question 1.

If C_{0}, C_{1}, C_{2},… C_{n} are coefficients of expansion (1 + x)^{n} then find the value of :

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

^{8}C_{1}+ ^{8}C_{3} + ^{8}C_{5} + ^{8}C_{7
}Solution:

(i) ^{8}C_{1} + ^{8}C_{2} + ^{8}C_{3} + … + ^{8}C_{8} We know that

^{n}C_{0} + ^{n}C_{1}+ ^{n}C_{2} + … + ^{n}C_{n} = 2^{n
}Putting n = 8

^{8}C_{0} + ^{8}C_{1} + ^{8}C_{2} + …. + ^{8}C_{8} = 2^{8
}⇒ 1 +^{8}C_{1} + ^{8}C_{2} + ^{8}C_{3} + … + ^{8}C_{8} = 2^{8
}⇒ ^{8}C_{1} + ^{8}C_{2} + …. + ^{8}C_{8} = 2^{8} – 1 = 255.

(ii) ^{8}C_{1} + ^{8}C_{3} + ^{8}C_{5} + ^{8}C_{7}_{
}We know that

^{8}C_{1} + ^{8}C_{3} + ^{8}C_{5} +… = 2^{n-1}

Putting n = 8

^{8}C_{1} + ^{8}C_{3} + ^{8}C_{5} + ^{8}C_{7} = 2^{8} ^{-1} = 2^{7} = 128.

Question 2.

If C_{0}, C_{1}, C_{2},… C_{n} are coefficients of expansion (1 + x)^{n }then prove that:

C_{0} + 3.C_{1} + 5.C_{2} + … + (2n + 1). C_{n} = (n + 1)2^{n}.

Solution:

L.H.S.

= ^{n}C_{0} + 3.C_{1} + 5.C_{2} + … + (2n + 1) ^{n}C_{n}

= ^{n}C_{0} + (2 + 1) ^{n}C_{1} + (4 + 1) ^{n}C_{2} + … (2n + 1) ^{n}C_{n
}= ^{n}C_{0} + (2. ^{n}C_{1} + ^{n}C_{1} + (4. ^{n}C_{2} + ^{n}C_{2}) + … (2n.^{n}C_{n} + ^{n}C_{n})

= 2. ^{n}C_{1} + 4 ^{n}C_{2} + 6 ^{n}C_{3} + … + 2n ^{n}C_{n}

+ (^{n}C_{0} + ^{n}C_{1} + ^{n}C_{2} + ^{n}C_{3} + …^{n}C_{n})

= 2[^{n}C_{1} + 2 ^{n}C_{2} + 3.^{n}C_{3} +… + ^{n}C_{n}] + (1 + 1)^{n}

Question 3.

C_{0 }C_{2} + C_{1 }C_{3} + C_{2 }C_{4} + … + C_{n} _ _{2}C_{n}= \(\frac { (2n)! }{ (n-2)!(n+2)! }\)

Solution:

From Binomial theorem

(1 + x)^{n} = C_{0} + C_{1}x + C_{2}x^{2} + … + C_{r}x^{r} + … + C_{n}x^{n} … (i)

Again

(x + 1)^{n} = C_{0}x^{n} + C_{1}x^{n-}^{1} + C_{2}x^{n-}^{2} + …+ C_{r}x^{n–}^{r}+ …+ C_{n-1 }x + C_{n} …(ii)

Multiplying equation (i) and (ii) we get

(1 + x)^{2n} = [C_{0} + C_{1}x + C_{2}x^{2} + … + C_{r}x^{r} + … + C_{n}x^{n}]

× [C_{0}x^{n} + C_{1}x^{n-}^{1} + C_{2}x^{n-}^{2} + …+ C_{r}x^{n–}^{r}+ …+ C_{n-1 }x + C_{n}]

Comparing cofficient of x^{n–}^{r} in both sides, we get

C_{0}C_{r} + C_{1}C_{r}+_{1}+ C_{2}C_{r} +_{2} + … + C_{n} _ _{r}C_{n
}

Question 4.

C_{0} + 2C_{1} + 4C_{2} + 6C_{3} + … + 2nC_{n} = 1 + n2^{n}.

Solution:

L.H.S.

= C_{0} + 2C_{1} + 4C_{2} + 6C_{3} + … + 2nC_{n
}= C_{0} + 2[C_{1} + 2C_{2} + 3C_{3} + … + nC_{n}]

Question 5.

Solution:

L.H.S.

Question 6.

If expansion of (1 + x – 2x^{2})^{6} is denoted by 1 + a_{1}x+ a_{2}x^{2} + a_{3}x^{3} + … + a_{12}x^{12 }then prove that a_{2} + a_{4} + a_{6} + … + a_{12 }= 31.

Solution :

Given expansion is :

(1 + x – 2x^{2})^{6} = 1 + a_{1}x+ a_{2}x^{2} + a_{3}x^{3} + … + a_{12}x^{12}^{ }… (i)

Putting x = – 1 in equation (i)

{1 + 1 – 2(1)^{2}}^{6} = 1+ a_{1} + a_{2} + a_{3} + … + a_{12}

⇒ (2 – 2)^{6} = 1 + a_{1} + a_{2} + a_{3} + … + a_{12
}⇒ 1 + a_{1} + a_{2} + a_{3} + … + a_{12 }= 0 … (ii)

Putting x = – 1 in equation (i)

{(1 – 1 – 2(-1)^{2}}^{6} = 1 – a_{1} + a_{2} + a_{3} + … + a_{12
}⇒ (-2)^{6} = 1 – a_{1} + a_{2} + a_{3} + … + a_{12}

⇒ 1- a_{1} + a_{2} + a_{3} + … + a_{12} = 64 … (iii)

Adding equation (ii) and (iii) we get

⇒ 2 + 2a_{2} + 2a_{4} + … + a_{12} = 64

⇒ 2(1 + a_{2} + a_{4} + … + a_{12} = 64

⇒ 1 + a_{2} + a_{4} + … + a_{12} = 64/2 =32

⇒ a_{2} + a_{4} + a_{6} … + a_{12} = 32 – 1 = 31

Hence, a_{2} + a_{4} + a_{6} … + a_{12} =31.

Hence proved.

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