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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)`
Solution
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 Tr+1 = nCr
a2 = coefficient of Tr+2 = nCr+1
a3 = coefficient of Tr+3 = nCr+2
And a4 = coefficient of Tr+4 = nCr+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)`
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