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Binomial Theorem

Short Answer Type

171. The coefficients of the (r - 1)^th, r^th and (r + 1)^th terms in the expansion of (x + 1)ⁿ are in the ratio 1 : 3 : 5. Find n and r.

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Long Answer Type

172.

Find the absolute term in the expansion of: $open parentheses straight x minus 1 over straight x squared close parentheses to the power of 3 straight n end exponent$

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173. The co-efficients of a^r-1, a^r and a^r+1 in the expansion of (1 + a)ⁿ are in arithmetic progression, Show that n² - n(4r + 1) + 4r² - 2 =0.

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Multiple Choice Questions

174.

The sum of coefficients of integral powers of x in the binomial expansion of $left parenthesis 1 minus 2 square root of straight x right parenthesis to the power of 50 space$ is

$1 half left parenthesis 3 to the power of 50 space plus space 1 right parenthesis$
$1 half left parenthesis 30 to the power of 50 right parenthesis$
$1 half left parenthesis 30 to the power of 50 minus 1 right parenthesis$
$1 half left parenthesis 30 to the power of 50 minus 1 right parenthesis$

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175.

If (10)⁹ +2 (11)²(10)⁷ + .....+10 (11)⁹ = K(10)⁹, then k is equal to

121/10
441/100
100
100

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176.
A multiple choice examination has 5 questions. Each question has three alternative answers of which exactly one is correct. The probability that a student will get 4 or more correct answers just by guessing is

17/3⁵

13/3⁵

11/3⁵

11/3⁵

11/3⁵

straight C presuperscript 5 subscript 4 space open parentheses 1 third close parentheses to the power of 4.2 over 3 space plus space straight C presuperscript 5 subscript 5 open parentheses 1 third close parentheses to the power of 5 left square bracket because space using space Binomial space distribution space to the power of straight n straight C subscript straight r straight p to the power of straight r straight q to the power of straight n minus straight r end exponent right square bracket Here space space straight n equals space 5 comma space straight r space equals space 0 comma 1 comma 2 comma 3 comma 4 comma 5 space equals space 5.2 over 3 to the power of 5 plus 1 over 3 to the power of 5 space equals space 11 over 3 to the power of 5

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177.

The term independent of x in $open parentheses fraction numerator straight x plus 1 over denominator straight x to the power of 2 divided by 3 end exponent minus straight x to the power of 1 divided by 3 end exponent space plus 1 end fraction minus fraction numerator straight x minus 1 over denominator straight x minus straight x to the power of 1 divided by 2 end exponent end fraction close parentheses to the power of 10$ is

171 Views

178.

If n is a positive integer, then $open parentheses square root of 3 plus 1 close parentheses to the power of 2 straight n end exponent minus space left parenthesis square root of 3 minus 1 right parenthesis to the power of 2 straight n end exponent space$ is