- Home
- Standard 11
- Mathematics
7.Binomial Theorem
hard
Sum of odd terms is $A$ and sum of even terms is $B$ in the expansion ${(x + a)^n},$ then
A
$AB = \frac{1}{4}{(x - a)^{2n}} - {(x + a)^{2n}}$
B
$2AB = {(x + a)^{2n}} - {(x - a)^{2n}}$
C
$4AB = {(x + a)^{2n}} - {(x - a)^{2n}}$
D
None of these
Solution
(c)${(x + a)^n} = {\,^n}{C_0}{x^n} + {^n}{C_1}{x^{n – 1}}a + {\,^n}{C_2}{x^{n – 2}}{a^2} + {\,^n}{C_3}{x^{n – 3}}{a^3} + …..$
But by the condition,
$A = {\,^n}{C_0}{x^n} + {\,^n}{C_2}{x^{n – 2}}{a^2} + {\,^n}{C_4}{x^{n – 4}}{a^4} + ……$
and $B = {\,^n}{C_1}{x^{n – 1}}a + {\,^n}{C_3}{x^{n – 3}}{a^3} + ……$
Hence $AB = \frac{1}{4}\left\{ {{{(x + a)}^{2n}} – {{(x – a)}^{2n}}} \right\}$
or $4AB = {(x + a)^{2n}} – {(x – a)^{2n}}$
Standard 11
Mathematics
