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