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7.Binomial Theorem
medium
If the coefficient of the middle term in the expansion of ${(1 + x)^{2n + 2}}$ is $p$ and the coefficients of middle terms in the expansion of ${(1 + x)^{2n + 1}}$ are $q$ and $r$, then
A
$p + q = r$
B
$p + r = q$
C
$p = q + r$
D
$p + q + r = 0$
Solution
(c) Since $(n+2)^{th}$ term is the middle term in the expansion of ${(1 + x)^{2n + 2}}$, therefore $p = {\,^{2n + 2}}{C_{n + 1}}$.
Since $(n+1)^{th}$ and $(n+2)^{th}$ terms are middle terms in the expansion of $(1+x)^{2n+1}$, therefore $q = {\,^{2n + 1}}{C_n}$ and $r = {\,^{2n + 1}}{C_{n + 1}}$ But $^{2n + 1}{C_n} + {\,^{2n + 1}}{C_{n + 1}} = {\,^{2n + 2}}{C_{n + 1}}$
$\therefore \,\,\,q + r = p$
Standard 11
Mathematics
