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7.Binomial Theorem
hard
${(1 + x)^n}$ ની વિસ્તરણમાં $p^{th}$ અને ${(p + 1)^{th}}$ પદના સહગુણક અનુક્રમે $p$ અને $q$ હોય તો $p + q = $
A
$n + 3$
B
$n + 1$
C
$n + 2$
D
$n$
Solution
(b) term $ = {T_p} = {}^n{C_{p – 1}}{(x)^{n – p + 1}}{(1)^{p – 1}} = p$
$(p + 1)^{th}$ term $ = {T_{p + 1}} = {}^n{C_p}{(x)^{n – p}}{(1)^p} = q$
Then, coefficient of $\frac{p}{q} = \frac{{{}^n{C_{p – 1}}}}{{{}^n{C_p}}}$
==> $\frac{p}{q} = \frac{{n!}}{{\left( {p – 1} \right)\,!\,\left( {n – p + 1}\right)\,\,!}}\,.\,\frac{{p\,!\,\,\,\left( {n – p} \right)\,\,!}}{{n\,!}}$
==> $\frac{p}{q} = \frac{p}{{n – p + 1}}$
==> $p + q = n + 1$.
Standard 11
Mathematics
