Question

If the three consecutive in the expansion of `(1+x)^n` are 28, 56, and 70, then the value of `n` is.

Answer 1

Correct Answer - 8
Let the three cosecutive coefficient be `.^(n)C_(r-1)=28,.^(n)C_(r)=56` and `.^(n)C_(r+1)=70,`
so that `(.^(n)C_(r))/(.^(n)C_(r-1))=(n-r+1)/ ( r)=(56)/(28)=2` and `(.^(n)C_(r+1))/(.^(n)C_(r))=(n-r)/(r+1)=(70)/(56)=(5)/(4)`
This gives n+1 =3r and 4n-5=9r. Therefore, `(4n-5)/(n+1)=3 or n=8`