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If the coefficients of three consecutive...

If the coefficients of three consecutive terms in the expansion of `(1+x)^n` are in the ratio 1:7:42, then find the value of `ndot`

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Let `(r+1)` th, `(r+2)`t h, and `(r+3)`th be the three consecutive terms. Then,
`.^(n)C_(r): .^(n)C_(r+1)=1:7:42`
Now, `(.^(n)C_(r))/(.^(n)C_(r+1))=1/7` or `(r+1)/(n-r)=1/7` or `n-8r = 7" "(1)`
and `(.^(n)C_(r+1))/(.^(n)C_(r+2))=7/42` or `(r+2)/(n-r-1)=1/6`
`rArr n-7r=13" "(2)`
Solving (1) and (2), we get `n = 55`.
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