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Exercise 5.1 · Q15

Q.In the binomial coefficients of (1+x)n(1+x)^n, the coefficients of the 5th5^{th}, 6th6^{th} and 7th7^{th} terms are in AP. Find all values of nn.

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Write the AP condition on nC4,nC5,nC6{}^nC_4,{}^nC_5,{}^nC_6 (the coefficients of the 5th5^{th}, 6th6^{th}, 7th7^{th} terms), divide by nC5{}^nC_5, and use consecutive-coefficient ratios to get a quadratic in nn.

Step 1. Identify the coefficients. T5=nC4, T6=nC5, T7=nC6T_5={}^nC_4,\ T_6={}^nC_5,\ T_7={}^nC_6. AP means 2 nC5=nC4+nC62\,{}^nC_5={}^nC_4+{}^nC_6.

Step 2. Divide through by nC5{}^nC_5. 2=nC4nC5+nC6nC52 = \dfrac{{}^nC_4}{{}^nC_5}+\dfrac{{}^nC_6}{{}^nC_5}.

Step 3. Use the ratio identities. nC4nC5=5n−4\dfrac{{}^nC_4}{{}^nC_5}=\dfrac5{n-4} and nC6nC5=n−56\dfrac{{}^nC_6}{{}^nC_5}=\dfrac{n-5}6. So

2=5n−4+n−56.2 = \frac5{n-4}+\frac{n-5}6.

Step 4. Clear denominators (multiply by 6(n−4)6(n-4)).

12(n−4)=30+(n−5)(n−4)12(n-4) = 30+(n-5)(n-4)

12n−48=30+n2−9n+2012n-48 = 30+n^2-9n+20

12n−48=n2−9n+5012n-48 = n^2-9n+50

0=n2−21n+980 = n^2-21n+98. …

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