Using binomial theorem, expand each of the following:

Question:

Using binomial theorem, expand each of the following:

$(3 x+2 y)^{5}$

 

Solution:

To find: Expansion of $(3 x+2 y)^{5}$

Formula used: (i) ${ }^{n} C_{r}=\frac{n !}{(n-r) !(r) !}$

(ii) $(a+b)^{n}={ }^{n} C_{0} a^{n}+{ }^{n} C_{1} a^{n-1} b+{ }^{n} C_{2} a^{n-2} b^{2}+\ldots \ldots+{ }^{n} C_{n-1} a b^{n-1}+{ }^{n} C_{n} b^{n}$

We have, $(3 x+2 y)^{5}$

$\Rightarrow\left[{ }^{5} C_{0}(3 x)^{5-0}\right]+\left[{ }^{5} C_{1}(3 x)^{5-1}(2 y)^{1}\right]+\left[{ }^{5} C_{2}(3 x)^{5-2}(2 y)^{2}\right]+\left[{ }^{5} C_{3}(3 x)^{5-3}(2 y)^{3}\right]+\left[{ }^{5} C_{4}(3 x)^{5-}\right.$

$\left.{ }^{4}(2 y)^{4}\right]+\left[{ }^{5} C_{5}(2 y)^{5}\right]$

$\Rightarrow\left[\frac{5 !}{0 !(5-0) !}\left(243 x^{5}\right)\right]+\left[\frac{5 !}{1 !(5-1) !}\left(81 x^{4}\right)(2 y)\right]+$

$\left[\frac{5 !}{2 !(5-2) !}\left(27 x^{3}\right)\left(4 y^{2}\right)\right]+\left[\frac{5 !}{3 !(5-3) !}\left(9 x^{2}\right)\left(8 y^{3}\right)\right]+$

$\left[\frac{5 !}{4 !(5-4) !}(3 x)\left(16 y^{4}\right)\right]+\left[\frac{5 !}{5 !(5-5) !}\left(32 y^{5}\right)\right]$

$\Rightarrow\left[1\left(243 x^{5}\right)\right]+\left[5\left(81 x^{4}\right)(2 y)\right]+\left[10\left(27 x^{3}\right)\left(4 y^{2}\right)\right]+\left[10\left(9 x^{2}\right)\left(8 y^{3}\right)\right]+\left[5(3 x)\left(16 y^{4}\right)\right]+$

$\Rightarrow 243 x^{5}+810 x^{4} y+1080 x^{3} y^{2}+720 x^{2} y^{3}+240 x y^{4}+32 y^{5}$

Ans) $243 x^{5}+810 x^{4} y+1080 x^{3} y^{2}+720 x^{2} y^{3}+240 x y^{4}+32 y^{5}$

 

Leave a comment

Close
faculty

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now