If the coefficient of (r-5)th and (2r-1)th term in the expansion of

Q) If the coefficient of (r-5)^th and (2r-1)^thterm in the expansion of (1+x)³⁴ are equal, then find the value of r.

[Practice Paper 1, 2023-24, Dir of Edu, GNCT of Delhi]

Ans:

Step 1: Since in binomial expansion of (x + y)ⁿ, the term T_r+1 is given by:

T_r+1 = $^nC_r$ X_n-r Y_r

Let’s start with (r-5)^th term:

T_r-5 = $^3^4C_{r-6} (1)^{34-(r-6)} (x)^{r-6}$

∴ T_r-5 = $^3^4C_{r-6} (x)^{r-6}$ (∵ value of any term with any power of 1 is always 1)

Similarly, (2r-1)^th term is given by:

T_2r-1 = $^3^4C_{2r-2} (1)^{34-(2r-2)} (x)^{2r-2}$

∴ T_2r-1 = $^3^4C_{2r-2} (x)^{2r-2}$

Step 2: Since it is given that coefficients of (r-5)^th and (2r-1)^th term are equal

∴ $^3^4C_{2r-2} = ^{34}C_{r-6}$

∴ $\frac{34!}{(2r-2)!(34-(2r-2))!} = \frac{34!}{(r-6)!(34-(r-6))!}$

∴ (2r-2)! (34-(2r-2))! = (r-6)! (34-(r-6))!

Step 3: To make the equation balanced, its terms of LHS should be equal to RHS.

Let’s take them one by one:

Let’s take: 2r – 2 = r – 6

∴ 2r – r = – 6 + 2

∴ r = – 4 but this is not possible

Step 4: Hence, let’s take: 2r – 2 = 34-(r-6)

∴ 2r – 2 = 34 – r + 6

∴ 2r + r = 34 + 6 + 2

∴ 3r = 42

∴ r = 14

Therefore, the value of r is 14.

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