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Miscellaneous · Q26

Q.If the coefficients of the 22nd, 33rd and 44th terms in the expansion of (1+x)n(1+x)^n are in arithmetic progression, find the value of nn.

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The 22nd, 33rd and 44th terms of (1+x)n(1+x)^n have coefficients nC1=n^{n}C_1=n, nC2=n(n−1)2^{n}C_2=\dfrac{n(n-1)}{2}, nC3=n(n−1)(n−2)6^{n}C_3=\dfrac{n(n-1)(n-2)}{6}. The A.P. condition is 2⋅nC2=nC1+nC32\cdot{}^{n}C_2={}^{n}C_1+{}^{n}C_3: n(n−1)=n+n(n−1)(n−2)6n(n-1)=n+\dfrac{n(n-1)(n-2)}{6}. Since n≠0n\ne0, divide through by nn: (n−1)=1+(n−1)(n−2)6(n-1)=1+\dfrac{(n-1)(n-2)}{6}. Multiply by 66: 6(n−1)=6+(n−1)(n−2)6(n-1)=6+(n-1)(n-2), i.e. 6n−6=6+n2−3n+26n-6=6+n^2-3n+2, so n2−9n+14=0n^2-9n+14=0, giving (n−7)(n−2)=0(n-7)(n-2)=0, i.e. n=7n=7 or n=2n=2. But n=2n=2 gives only 33 terms in the expansion, so there is no 44th term -- n=2n=2 is rejected. [!ANSWER] n=7n=7.

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