Menu Close

prove-that-1-x-n-1-nx-n-n-1-2-x-2-n-n-1-n-2-3-x-3-n-n-n-using-a-suitable-expansion-method-hence-determine-the-expansion-of-2-001-89-




Question Number 55815 by Rio Mike last updated on 04/Mar/19
prove that  (1+x)^n = 1+nx +((n(n−1))/(2!))x^2 +((n(n−1)(n−2))/(3!))x^3 +...n(n−n)  using a suitable expansion method  hence determine the expansion of  (2.001)^(89)
$${prove}\:{that} \\ $$$$\left(\mathrm{1}+{x}\right)^{{n}} =\:\mathrm{1}+{nx}\:+\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}!}{x}^{\mathrm{2}} +\frac{{n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)}{\mathrm{3}!}{x}^{\mathrm{3}} +…{n}\left({n}−{n}\right) \\ $$$${using}\:{a}\:{suitable}\:{expansion}\:{method} \\ $$$${hence}\:{determine}\:{the}\:{expansion}\:{of} \\ $$$$\left(\mathrm{2}.\mathrm{001}\right)^{\mathrm{89}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Answered by Kunal12588 last updated on 04/Mar/19
we can expand (1+x)^n  using binomial  expansion.  (a+b)^n =^n C_0 a^n +^n C_1 a^(n−1) b^1 +...+^n C_r a^(n−r) b^r +...+^n C_n b^n   put a=1 and b=x  (1+x)^n =^n C_0 1^n +^n C_1 1^(n−1) x+...+^n C_n x^n   =1+((n(n−1)!)/((n−1)!))x+((n(n−1)(n−2)!)/(2!(n−2)!))x^2 +...+x^n   =1+nx+((n(n−1))/(2!))x^2 +((n(n−1)(n−2))/(3!))x^3 +...+x^n   (2.001)^(89) =(1+1.001)^(89)   =1+89(1.001)+((89×88)/2)(1.001)^2 +...+(1.001)^(89)
$${we}\:{can}\:{expand}\:\left(\mathrm{1}+{x}\right)^{{n}} \:{using}\:{binomial} \\ $$$${expansion}. \\ $$$$\left({a}+{b}\right)^{{n}} =^{{n}} {C}_{\mathrm{0}} {a}^{{n}} +^{{n}} {C}_{\mathrm{1}} {a}^{{n}−\mathrm{1}} {b}^{\mathrm{1}} +…+^{{n}} {C}_{{r}} {a}^{{n}−{r}} {b}^{{r}} +…+^{{n}} {C}_{{n}} {b}^{{n}} \\ $$$${put}\:{a}=\mathrm{1}\:{and}\:{b}={x} \\ $$$$\left(\mathrm{1}+{x}\right)^{{n}} =^{{n}} {C}_{\mathrm{0}} \mathrm{1}^{{n}} +^{{n}} {C}_{\mathrm{1}} \mathrm{1}^{{n}−\mathrm{1}} {x}+…+^{{n}} {C}_{{n}} {x}^{{n}} \\ $$$$=\mathrm{1}+\frac{{n}\left({n}−\mathrm{1}\right)!}{\left({n}−\mathrm{1}\right)!}{x}+\frac{{n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)!}{\mathrm{2}!\left({n}−\mathrm{2}\right)!}{x}^{\mathrm{2}} +…+{x}^{{n}} \\ $$$$=\mathrm{1}+{nx}+\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}!}{x}^{\mathrm{2}} +\frac{{n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)}{\mathrm{3}!}{x}^{\mathrm{3}} +…+{x}^{{n}} \\ $$$$\left(\mathrm{2}.\mathrm{001}\right)^{\mathrm{89}} =\left(\mathrm{1}+\mathrm{1}.\mathrm{001}\right)^{\mathrm{89}} \\ $$$$=\mathrm{1}+\mathrm{89}\left(\mathrm{1}.\mathrm{001}\right)+\frac{\mathrm{89}×\mathrm{88}}{\mathrm{2}}\left(\mathrm{1}.\mathrm{001}\right)^{\mathrm{2}} +…+\left(\mathrm{1}.\mathrm{001}\right)^{\mathrm{89}} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *