if three successive coefficients in the expressions of (1+x)n are 220, 495 and 792 respectively, find the value of n?

LetTr , Tr+1 , Tr+2 be the sucessive terms in the expansionof (1+x)n

Now Tp+1 in  (1+x)n = nCpxp    => coefficient of (p+1)th term = nCp

. :  Coefficient of rth term = nCr-1 = 220 ..........(1)    { p+1 = r    =>  p = r-1 }

      Coefficient of (r+1)th term = nCr =495 ..........(2)    { p+1 = r+1 =>  p = r}

      Coefficient of (r+2)th term = nCr+1 = 792 ..........(3)    { p+1 = r+

  • -5

 LetTr , Tr+1 , Tr+2 be the sucessive terms in the expansionof (1+x)n

Now Tp+1 in  (1+x)n = nCpxp   => coefficient of (p+1)th term = nCp

. :  Coefficient of rth term = nCr-1 = 220 ..........(1)    { p+1 = r    =>  p = r-1 }

      Coefficient of (r+1)th term = nCr =495 ..........(2)    { p+1 = r+1 =>  p = r}

      Coefficient of (r+2)th term = nCr+1 = 792 ..........(3)    { p+1 = r+2   =>  p = r+1 }

Dividing (2) by (1) we get

nC/ nCr-1 = 495 / 220        

 => n! / (n-r) ! r ! * (n-r+1) ! (r-1) ! / n !  

=> n - r +1 / r = 9/4

=> 4(n - r +1)  = 9r

=> 4n - 4r +4  = 9r 

=> 4n +4 = 9r + 4r

=>  4n - 13r +4 = 0 ...........(4)

Dividing (3) by (2) we get

nCr+1 / nCr  = 792/ 495

=> n! / (n-r-1) ! (r+1) ! * (n-r) ! r ! / n !   = 8/5

=> n - r / r + 1 = 8 / 5

=> 5n - 13r -8 = 0

Subtracting (5) from (4) we get 

  4n - 13r +4 = 0 

   5n - 13r -8 = 0                                                                                                                                                     &nb

  • 37

LetTr , Tr+1 , Tr+2 be the sucessive terms in the expansionof (1+x)n

Now Tp+1 in  (1+x)n = nCpxp   => coefficient of (p+1)th term = nCp

. :  Coefficient of rth term = nCr-1 = 220 ..........(1)    { p+1 = r    =>  p = r-1 }

      Coefficient of (r+1)th term = nCr =495 ..........(2)    { p+1 = r+1 =>  p = r}

      Coefficient of (r+2)th term = nCr+1 = 792 ..........(3)    { p+1 = r+2   =>  p = r+1 }

Dividing (2) by (1) we get

nC/ nCr-1 = 495 / 220        

=> n - r +1 / r = 9/4

=> 4(n - r +1)  = 9r

=> 4n - 4r +4  = 9r 

=> 4n +4 = 9r + 4r

=>  4n - 13r +4 = 0 ...........(4)

Dividing (3) by (2) we get

nCr+1 / nCr  = 792/ 495

=> n - r / r + 1 = 8 / 5

=> 5n - 13r -8 = 0

Subtracting (5) from (4) we get 

  4n - 13r +4 = 0 

   5n - 13r -8 = 0                                                                                                                                                                                               %2

  • -1

LetTr , Tr+1 , Tr+2 be the sucessive terms in the expansionof (1+x)n

Now Tp+1

  • 1

thank you so much @geethanjali!!!!! :)

  • -3

LetTr , Tr+1 , Tr+2 be the sucessive terms in the expansionof (1+x)n

Now Tp+1 in (1+x)n = nCpxp => coefficient of (p+1)th term = nCp

. : Coefficient of rth term = nCr-1 = 220 ..........(1) { p+1 = r => p = r-1 }

Coefficient of (r+1)th term = nCr =495 ..........(2) { p+1 = r+1 => p = r}

Coefficient of (r+2)th term = nCr+1 = 792 ..........(3) { p+1 = r

  • 6

LetTr , Tr+1 , Tr+2 be the sucessive terms in the expansionof (1+x)n

Now Tp+1 in (1+x)n = nCpxp => coefficient of (p+1)th term = nCp

. : Coefficient of rth term = nCr-1 = 220 ..........(1) { p+1 = r => p = r-1 }

Coefficient of (r+1)th term = nCr =495 ..........(2) { p+1 = r+1 => p = r}

Coefficient of (r+2)th term = nCr+1 = 792 ..........(3) { p+1 = r

  • -2

LetTr , Tr+1 , Tr+2 be the sucessive terms in the expansionof (1+x)n

Now Tp+1 in (1+x)n = nCpxp => coefficient of (p+1)th term = nCp

. : Coefficient of rth term = nCr-1 = 220 ..........(1) { p+1 = r => p = r-1 }

Coefficient of (r+1)th term = nCr =495 ..........(2) { p+1 = r+1 => p = r}

Coefficient of (r+2)th term = nCr+1 = 792 ..........(3) { p+1 = r

  • -1

LetTr , Tr+1 , Tr+2 be the sucessive terms in the expansionof (1+x)n

Now Tp+1 in (1+x)n = nCpxp => coefficient of (p+1)th term = nCp

. : Coefficient of rth term = nCr-1 = 220 ..........(1) { p+1 = r => p = r-1 }

Coefficient of (r+1)th term = nCr =495 ..........(2) { p+1 = r+1 => p = r}

Coefficient of (r+2)th term = nCr+1 = 792 ..........(3) { p+1 = r

  • -5

LetTr , Tr+1 , Tr+2 be the sucessive terms in the expansionof (1+x)n

Now Tp+1 in (1+x)n = nCpx=> coefficient of (p+1)th term = nCp

. : Coefficient of rth term = nCr-1 = 220 ..........(1) { p+1 = r => p = r-1 }

Coefficient of (r+1)th term = nCr =495 ..........(2) { p+1 = r+1 => p = r}

Coefficient of (r+2)th term = nCr+1 = 792 ..........(3) { p+1 = r+2 => p = r+1 }

Dividing (2) by (1) we get

nC/ nCr-1 = 495 / 220

=> n - r +1 / r = 9/4

=> 4(n - r +1) = 9r

=> 4n - 4r +4 = 9r

=> 4n +4 = 9r + 4r

=> 4n - 13r +4 = 0 ...........(4)

Dividing (3) by (2) we get

nCr+1 / nC= 792/ 495

=> n - r / r + 1 = 8 / 5

=> 5n - 13r -8 = 0

Subtracting (5) from (4) we get

4n - 13r +4 = 0

5n - 13r -8 = 0 %2

hope this helps you!!!

  • 14
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