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Question Number 37728 by kunal1234523 last updated on 17/Jun/18

$$1+\mathrm{n}+\frac{\mathrm{n}(\mathrm{n}-1)}{2!}+\frac{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)}{3!}+\frac{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)(\mathrm{n}-3)}{4!}+........=$$

Answered by tanmay.chaudhury50@gmail.com last updated on 17/Jun/18

$${(1+x)}^n=1+{nC}_{1}x+{nC}_2{x}^2+{nC}_3{x}^3+...+{nC}_n{x}^n$$$$putx=1$$$$2^n=1+{nC}_1+{nC}_2+{nC}_3+...+{nC}_n$$$$2^n=1+n+\frac{n(n-1)}{2!}+\frac{n(n-1)(n-2)}{3!}$$$$sotheansweris{2}^n$$

Commented by kunal1234523 last updated on 17/Jun/18

Commented by kunal1234523 last updated on 17/Jun/18

$$\mathrm{according}\mathrm{to}\mathrm{wolfram}\mathrm{alpha}\mathrm{it}\mathrm{can}\mathrm{not}\mathrm{be}$$$$\mathrm{determine}\mathrm{in}\mathrm{general}\mathrm{form}\mathrm{but}\mathrm{you}\mathrm{did}\mathrm{it}$$$$\mathrm{are}\mathrm{you}\mathrm{a}\mathrm{genius}$$

Commented by kunal1234523 last updated on 17/Jun/18

$$\mathrm{This}\mathrm{question}\mathrm{is}\mathrm{arived}\mathrm{to}\mathrm{me}\mathrm{when}\mathrm{I}\mathrm{am}$$$$\mathrm{working}\mathrm{on}\mathrm{sets}.\mathrm{I}\mathrm{came}\mathrm{to}\mathrm{know}\mathrm{that}\mathrm{no}.\mathrm{of}$$$$\mathrm{subsets}\mathrm{of}\mathrm{a}\mathrm{set}\mathrm{is}{2}^{\mathrm{n}}\mathrm{where}\mathrm{n}\mathrm{is}\mathrm{no}.\mathrm{of}\mathrm{terms}$$$$\mathrm{or}\mathrm{objects}\mathrm{in}\mathrm{the}\mathrm{set}.\mathrm{so}\mathrm{I}\mathrm{tried}\mathrm{to}\mathrm{prove}\mathrm{it}\mathrm{and}$$$$\mathrm{I}\mathrm{end}\mathrm{up}\mathrm{in}\mathrm{the}\mathrm{question}\mathrm{that}\mathrm{I}\mathrm{had}\mathrm{asked}.$$

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