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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 n.
1. 45
2. 65
3. 55
4. 75

1 Answer

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Best answer
Correct Answer - Option 3 : 55

Calculation:

⇒ Let (r – 1)th, (r)th and (r + 1)th be the three consecutive terms.

Then, the given ratio is 1:7:42

Now (nCr-2 / nCr – 1) = (1/7)

⇒ (nCr-2 / nCr – 1) = (1/7) ⇒ [(r – 1)/(n − r+2)] = (1/7) ⇒ n − 8r + 9 = 0 → (1)

And,

⇒ (nCr-1 / nCr) = (7/42) ⇒ [(r)/(n – r +1)] =(1/6) ⇒ n − 7r   + 1 = 0 → (2)

From (1) & (2), n = 55

Ratio of Consecutive Terms/Coefficients:

⇒ Coefficients of xr and xr + 1 are nCr – 1 and nCr respectively.

⇒ (nCr / nCr – 1) = (n – r + 1) / r

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