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Binomial Theorem

Short Answer Type

51. How many terms do the expansion of the following have?

left parenthesis 1 minus 2 straight x plus straight x squared right parenthesis to the power of 7

113 Views

52. Expand using Binomial Theorem.

open parentheses 1 plus straight x over 2 minus straight x over 2 close parentheses to the power of 4 comma space straight x not equal to 0

128 Views

Long Answer Type

53. Find the co-efficient of x⁵ in the product (1 + 2x)⁶ (1 -x)⁷ using Binomial theorem.

117 Views

54. Find the co-efficient of a⁴ in the product (1 + 2a)⁴ (2 - a)⁵ using Binomial Theorem.

104 Views

Short Answer Type

55. Using Binomial Theorem, expand the following expression:

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162 Views

56. Using Binomial Theorem, expand the following expression:

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173 Views

57. Expand the following expression, using Binomial Theorem:

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110 Views

58. Expand the following expression, using Binomial Theorem:

left parenthesis 1 plus straight x plus straight x squared right parenthesis cubed

101 Views

59.

Find the co-efficient of $straight x squared$ in the expansion of $open parentheses 3 straight x minus 1 over straight x close parentheses to the power of 6.$

95 Views

60.
Find the term independent of x in the expansion of $open parentheses straight x squared plus 1 over straight x close parentheses to the power of 9.$

Let $straight T subscript straight r plus 1 end subscript$ be the term independent of x or absolute term.
$rightwards double arrow space straight T subscript straight r plus 1 end subscript$ contains $straight x to the power of 0.$

Now,    $straight T subscript straight r plus 1 end subscript space equals space straight C presuperscript 9 subscript straight r left parenthesis straight x squared right parenthesis to the power of 9 minus straight r end exponent. space open parentheses 1 over straight x close parentheses to the power of straight r$
Power of x = (18 - 2r) - r
= 18 - 3r
Power of x required = 0
$rightwards double arrow$ 18 - 3r = 0 $rightwards double arrow$ r = 6
$straight T subscript straight r plus 1 end subscript space equals space straight T subscript 6 plus 1 end subscript space equals space straight T subscript 7$ is independent of x.
Now,    $straight T subscript 7 space equals space straight C presuperscript 9 subscript 6 left parenthesis straight x squared right parenthesis to the power of 9 minus 6 end exponent. space open parentheses 1 over straight x close parentheses to the power of 6$
   $equals fraction numerator 9 factorial over denominator 6 factorial space 3 factorial end fraction equals fraction numerator 9 cross times 8 cross times 7 over denominator 3 cross times 2 cross times 1 end fraction$
= 84
Hence, the absolute term is 84.