Maharashtra State Board 11th Maths Solutions Chapter 4 Methods of Induction and Binomial Theorem Ex 4.3
Question 1
Maharashtra Board Solution
In the following expansions, find the indicated term. (i) , 3rd term
Solution & Step-by-Step Answer:
(ii) , 5th term
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(iii) , 7th term
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(iv) In , 9th term
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(v) In , 10th term
Solution:
Question 2
Maharashtra Board Solution
In the following expansions, find the indicated coefficients. (i) x3 in
Solution & Step-by-Step Answer:
(ii) x8in
Solution:
(iii) x9in
Solution:
(iv) x-3in
Solution:
(v) x-20in
Solution:
Question 3
Maharashtra Board Solution
Find the constant term (term independent of x) in the expansion of (i)
Solution & Step-by-Step Answer:
(ii)
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(iii)
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(iv)
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(v)
Solution:
Question 4
Maharashtra Board Solution
Find the middle terms in the expansion of (i)
Solution & Step-by-Step Answer:
(ii)
Solution:
(iii)
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(iv)
Solution:
(v)
Solution:
Question 5
Maharashtra Board Solution
In the expansion of (k + x)8, the coefficient of x5 is 10 times the coefficient of x6. Find the value of k.
Solution & Step-by-Step Answer:
Question 6
Maharashtra Board Solution
Find the term containing x6 in the expansion of (2 – x) (3x + 1)9.
Solution & Step-by-Step Answer:
Question 7
Maharashtra Board Solution
The coefficient of x2 in the expansion of (1 + 2x)m is 112. Find m.
Solution & Step-by-Step Answer:
The coefficient of x2 in (1 + 2x)m = mC2 (22) Given that the coefficient of x2 = 112 ∴ mC2 (4) = 112 ∴ mC2 = 28 ∴ ∴ ∴ m(m – 1) = 56 ∴ m2 – m – 56 = 0 ∴ (m – 8) (m + 7) = 0 As m cannot be negative. ∴ m = 8