Binomial Theorem

More Topic from Mathematics

Question 1

Statement space minus 1 space colon space sum from straight r space equals space 0 to straight n of space left parenthesis straight r plus 1 right parenthesis space to the power of straight n straight C subscript straight r space equals space left parenthesis straight n plus 2 right parenthesis 2 to the power of straight n minus 1 end exponent
Statement space minus space 2 colon thin space sum from straight r equals 0 to straight n of space left parenthesis straight r plus 1 right parenthesis space to the power of straight n straight C subscript straight r straight x to the power of straight r space equals space left parenthesis 1 plus straight x right parenthesis to the power of straight n space plus space nx space left parenthesis 1 plus straight x right parenthesis to the power of straight n minus 1 end exponent
  • Statement −1 is false, Statement −2 is true

  • Statement −1 is true, Statement −2 is true, Statement −2 is a correct explanation for Statement −1

  • Statement −1 is true, Statement −2 is true; Statement −2 is not a correct explanation for Statement −1.

  • Statement − 1 is true, Statement − 2 is false. 

Solution

B.

Statement −1 is true, Statement −2 is true, Statement −2 is a correct explanation for Statement −1

sum from straight r space equals 0 to straight n of space left parenthesis straight r plus 1 right parenthesis space to the power of straight n straight C subscript straight r space equals space sum from straight r space equals space 0 to straight n of space straight r to the power of straight n straight C subscript straight r space plus space to the power of straight n straight C subscript straight r
space equals space sum from straight r space equals space 0 to straight n of space straight r space straight n over straight r space to the power of straight n minus 1 end exponent straight C subscript straight r minus 1 end subscript space plus space sum from straight r space equals space 0 to straight n of space to the power of straight n straight C subscript straight r space equals space straight n 2 to the power of straight n minus 1 end exponent space plus space 2 to the power of straight n
space equals space 2 to the power of straight n minus 1 end exponent space left parenthesis straight n plus 2 right parenthesis
Statement space minus 1 space true
sum for space of space left parenthesis straight r plus 1 right parenthesis to the power of straight n straight C subscript straight r space end subscript straight x to the power of straight r space equals space sum for space of straight r to the power of straight n space straight C subscript straight r straight x to the power of straight r space plus space sum for space of to the power of straight n straight C subscript straight r straight x to the power of straight r
space equals space straight n space sum from straight r space equals space 0 space to straight n of space to the power of straight n straight C subscript straight r minus 1 end subscript space straight x to the power of straight r space plus space sum from straight r space equals space 0 to straight n of space to the power of straight n straight C subscript straight r straight x to the power of straight r space equals space straight n space straight x space left parenthesis 1 space plus straight x right parenthesis to the power of straight n minus 1 end exponent space plus space left parenthesis 1 plus straight x right parenthesis to the power of straight n
substituting space straight x space equals space 1
sum for space of space left parenthesis straight r plus 1 right parenthesis space to the power of straight n straight C subscript straight r space equals space straight n 2 to the power of straight n minus 1 end exponent space plus space 2 to the power of straight n
Hence Statement −2 is also true and is a correct explanation of Statement −1.
Question 2

If space straight S subscript straight n space equals space sum from straight r equals 0 to straight n of fraction numerator 1 over denominator straight C presuperscript straight n subscript straight r end fraction space and space straight t subscript straight n space equals space sum from straight r equals space 0 to straight n of space straight r over straight C subscript straight r space then comma straight t subscript straight n over straight S subscript straight n space is
  • 1 half straight n
  • 1 half straight n minus 1
  • n-1

  • n

Solution

A.

1 half straight n straight S subscript straight n space equals space sum from straight r space equals 0 to straight n of fraction numerator 1 over denominator straight C presuperscript straight n subscript straight n minus straight r end subscript end fraction
straight S subscript straight n space equals space sum from straight r space equals 0 to straight n of fraction numerator 1 over denominator straight C presuperscript straight n subscript straight n minus straight r end subscript end fraction space left parenthesis because space to the power of straight n straight C subscript straight r space equals straight C presuperscript straight n subscript straight n minus straight r end subscript right parenthesis
ns subscript straight n space equals space sum from straight r space equals 0 to straight n of fraction numerator 1 over denominator straight C presuperscript straight n subscript straight n minus straight r end subscript end fraction space
ns subscript straight n space equals space sum from straight r space equals 0 to straight n of open square brackets fraction numerator straight n minus straight r over denominator straight C presuperscript straight n subscript straight n minus straight r end subscript end fraction plus fraction numerator straight r over denominator straight C presuperscript straight n subscript straight n minus straight r end subscript end fraction close square brackets
ns subscript straight n space equals space sum from straight r space equals 0 to straight n of space fraction numerator straight n minus straight r over denominator straight C presuperscript straight n subscript straight n minus straight r end subscript end fraction space plus space sum from straight r space equals 0 to straight n of space fraction numerator straight r over denominator straight C presuperscript straight n subscript straight r end fraction
ns subscript straight n space equals space straight t subscript straight n plus straight t subscript straight n
ns subscript straight n space equals space 2 straight t subscript straight n
straight t subscript straight n over straight s subscript straight n space equals straight n over 2
Question 3

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

  • 10/35

Solution

C.

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
Question 4

For natural numbers m, n if (1 − y)m (1 + y)n = 1 + a1y + a2y2 + … , and a1 = a2 = 10, then (m, n) is

  • (20, 45)

  • (35, 20)

  • (45, 35)

  • (35, 45)

Solution

D.

(35, 45)

left parenthesis 1 minus straight y right parenthesis to the power of straight m space left parenthesis 1 plus straight y right parenthesis to the power of straight n space equals space left square bracket 1 minus to the power of straight m straight C subscript 1 straight y space plus to the power of straight m straight C subscript 2 straight y squared space minus........ right square bracket left square bracket 1 plus to the power of straight n straight C subscript 1 straight y space plus to the power of straight n straight C subscript 2 straight y squared plus... right square bracket
space equals space 1 space plus space left parenthesis straight n minus straight m right parenthesis space plus open curly brackets fraction numerator straight m left parenthesis straight m minus 1 right parenthesis over denominator 2 end fraction plus fraction numerator straight n left parenthesis straight n minus 1 right parenthesis 2 over denominator 2 end fraction minus mn close curly brackets straight y squared space plus....
therefore space straight a subscript 1 space equals straight n minus straight m space equals space 10 space and space straight a subscript 2 space equals space fraction numerator straight m squared plus straight n squared space minus straight m minus straight n minus 2 mn over denominator 2 end fraction space equals space 10
So comma space straight n minus straight m equals 10 space and space left parenthesis straight m minus straight n right parenthesis squared space minus left parenthesis straight m plus straight n right parenthesis space equals space 20
rightwards double arrow straight m plus straight n space equals space 80
therefore comma space straight m space equals space 35 comma space straight n space equals space 45
Question 5

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

  • 121/10

  • 441/100

  • 100

  • 110

Solution

C.

100