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Question

The expansion of (1+x)n has 3 consecutive terms with coefficient in the ratio 1:2:3 and can be written in the form nCk;nCk+1:nCk+2.. The sum of all possible values of n+k is

A
18
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B
21
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C
28
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D
32
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Solution

The correct option is A 18
Given
nCk:nCk+1:nCk+2=1:2:3
nCknCk+1=12n!(nk1)(nk)!x(k)!×(k+1)!n!=dfrac12
(k+1)(nk)=12
2k+2=nk
n=3k+2------- (1)
nCk+1nCk+2=23n!(nk2)!(nk1)!(k+1)!23
(k+2)(nk1)=23
3k+6=2n2k2
5k+8=2n
On solving 6k+9=5k+8k=4n=x4+2
n+k=14+4=18
n=14

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