By expanding $[(1 + x)^{n} - 1]^{m}$ in two ways prove that $^{m} C_{1}^{n} C_{m} -^{m} C_{2}^{2 n} C_{m} +,^{m} C_{3}^{3 n} C_{m}, . . . = (- 1)^{m - 1} n^{m}$

Open in App

Solution

$[(1 + x)^{n} - 1]^{m} = [1 +^{n} C_{1} x +^{n} C_{2} x^{2} + - 1]^{m}$
$= x^{m} [^{n} C_{1} +^{n} C_{2} x +]^{m}$
$= x^{m} [n +^{n} C_{2} x + . .]^{m}$
coefficient of $x^{m} i n x^{n}$
Again the given expression can be written as
$(- 1)^{m} [1 - (1 + x)^{n}]^{m}$
$(- 1)^{m} [1 -^{m} C_{1} (1 + x)^{n} +^{m} C_{2} (1 + x)^{2 n} -^{m} C_{3} (1 + x)^{3 n}]$
Now we know that coefficient of $x^{r} i n (1 + x)^{n} i s^{n} C_{r}$ as it occurs
Hence coefficient of $x m$ in the above expansion of
$(- 1)^{m} [-^{m} C_{1}^{n} C_{m} +^{m} C_{2}^{2 n} C_{m} -^{m} C_{3}^{3 n} C_{m} +]$
Equating the coefficient of $x^{m}$ We get the result $(- 1)^{m} = - (1)^{m - 1}$ etc

Suggest Corrections

Similar questions

(1 + x)^{n} =^{n} C_{0} +^{n} C_{1} x +^{n} C_{2} x^{2} +^{n} C_{n} x^{n}, x ϵ N

^{n} C_{0} -^{n} C_{1} +^{n} C_{2} - . . . . . . + (- 1)^{m - 1} \times^{n} C_{m - 1}

= (- 1)^{m - 1} \frac{(n - 1) (n - 2) . . (n - m + 1)}{(m - 1)!}

Prove that : $(m + n)^{- 1} (m^{- 1} + n^{- 1}) = (m n)^{- 1}$

{(1 + x)}^{n} - n x - 1

x^{2}, n \in N

(x^{n} - y^{n})

(x - y) w h e r e x \neq y a n d n \in N .

x_{1}, x_{2}, x_{3}, . . . . . . . . . ., x_{n}

x_{n}

I = \int \frac{e^{x} (1 + n x^{n - 1} - x^{2 n})}{(1 - x^{n} \sqrt{1 - x^{2 n}})} d x = e^{x} \sqrt{\frac{1 + x^{n}}{1 - x^{n}}}

Join BYJU'S Learning Program