Question

If a₁, a₂, a₃ and a₄ are the coefficient of any four consecutive terms in the expansion of (1 + x)ⁿ, prove that `(a_1)/(a_1 + a_2) + (a_3)/(a_3 + a_4) = (2a_2)/(a_2 + a_3)`

Answer 1

Let a₁, a₂, a₃ and a₄ be the coefficient of four consecutive terms `"T"_(r + 1), "T"_(r + 2), "T"_(r + 3)` and `"T"_(r + 4)` respectively.

Then a₁ = coefficient of T_r+1 = ⁿC_r

a₂ = coefficient of T_r+2 = ⁿC_r+1 

a₃ = coefficient of T_r+3 = ⁿC_r+2

And a₄ = coefficient of T_r+4 = ⁿC_r+3

Thus `(a_1)/(a_1 + a_2) = (""^n"C"_r)/(""^n"C"_r + ""^n"C"_(r + 1))`

= `(""^n"C"_r)/(""^(n + 1)"C"_(r + 1)`   .....`(because ""^n"C"_r + ""^n"C"_(r + 1) = ""^(n + 1)"C"_(r + 1))`

= `(n)/(r(n - r)) xx  ((r + 1)(n - r))/(n + 1)`

= `(r + 1)/(n + 1)`

Similarly, `(a_3)/(a_3 + a_4) = (""^n"C"_(r + 2))/(""^n"C"_(r + 2) + ""^n"C"_(r + 3))`

= `(""^n"C"_(r + 2))/(""^(n + 1)"C"_(r + 3))`

= `(r + 3)/(n + 1)`

Hence, L.H.S. = `a_1/(a_1 + a_2) + a_3/(a_3 + a_4)`

= `(r + 1)/(n + 1) + (r + 3)/(n + 1)`

= `(2r + 4)/(n + 1)`

And R.H.S. = `(2a_2)/(a_2 + a_2) + a_3/(a_3 + a_4)`

= `(2(""^n"C"_(r + 1)))/(""^n"C"_(r + 1) + ""^n"C"_(r + 2))`

= `(2(""^n"C"_(r + 1)))/(""^(n + 1)"C"_(r + 2))`

= `2  n/((r + 1)(n - r - 1)) xx ((r + 2)(n - r - 1))/(n + 1)`

= `(2(r + 2))/(n + 1)`