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Question Number 120636 by TANMAY PANACEA last updated on 01/Nov/20

$$selectiveBinomial$$

Commented by TANMAY PANACEA last updated on 01/Nov/20

Commented by TANMAY PANACEA last updated on 01/Nov/20

Commented by TANMAY PANACEA last updated on 01/Nov/20

Commented by TANMAY PANACEA last updated on 01/Nov/20

Commented by TANMAY PANACEA last updated on 01/Nov/20

Commented by TANMAY PANACEA last updated on 01/Nov/20

Commented by TANMAY PANACEA last updated on 01/Nov/20

Commented by Dwaipayan Shikari last updated on 01/Nov/20

$$S=\frac{1}{1!(n-1)!}+\frac{1}{3!(n-3)!}+...$$$$S=\frac{n}{n!}+\frac{n(n-1)(n-2)}{3!n!}+\frac{n(n-1)(n-2)(n-3)(n-4)}{5!n!}+...$$$$=\frac{1}{n!}(n+\frac{n(n-1)(n-2)}{3!}+\frac{n(n-1)(n-2)(n-3)(n-4)}{5!}+...)$$$$=\frac{2^{n-1}}{n!}$$$${(1+x)}^n=1+nx+\frac{n(n-1)}{2!}{x}^2+...$$$${(1-x)}^n=1-nx+\frac{n(n-1)}{2!}{x}^2-..$$$$\frac{{(1+x)}^n-{(1-x)}^n}{2}=nx+\frac{n(n-1)(n-2)}{3!}{x}^3+...$$$$2^{n-1}=(n+\frac{n(n-1)(n-2)}{3!}+...)(x=1)$$

Commented by Dwaipayan Shikari last updated on 01/Nov/20

$${(1+x)}^n=C_0+C_{1}x+...$$$${\int }_0^1{(1+x)}^n={\int }_0^1{C}_0+C_{1}x+..$$$$\frac{2^{n+1}}{n+1}=C_0+\frac{1}{2}{C}_1+\frac{1}{3}{C}_2+...\frac{1}{n+1}{C}_n$$

Commented by Dwaipayan Shikari last updated on 01/Nov/20

$$C_0+(C_0+C_1)+(C_0+C_1+C_2)+...$$$$n(C_0+C_1+C_2+...)-(C_1+2C_2+3C_3+..)$$$$n.2^n-n.2^{n-1}=n{2}^{n-1}(2-1)=n.2^{n-1}$$$$[{(1+x)}^n=C_0+C_{1}x+C_3{x}^2+..$$$$n{(1+x)}^{n-1}=C_1+2C_{3}x+..$$$$n.2^{n-1}=C_1+2C_3+...]$$

Commented by Dwaipayan Shikari last updated on 01/Nov/20

$${(1+x+2x^2)}^{20}=a_0+a_{1}x+...$$$$1^{20}=a_0(x=0)$$$${\left(2\right)}^{40}=1+a_1+a_2+....(x=1)$$$$2^{20}=1-a_1+a_2-a_3+....(x=-1)$$$$2^{40}+2^{20}=2(1+a_2+a_4+...+a_{38})$$$$2^{39}+2^{19}=a_0+a_2+....$$$$2^{19}(2^{20}+1)=a_0+a_2+...+a_{38}$$

Commented by TANMAY PANACEA last updated on 01/Nov/20

$$thankyou$$

Commented by TANMAY PANACEA last updated on 01/Nov/20

$$ihavepostedquestionscollectedfrom[booksofIIT$$

Commented by Dwaipayan Shikari last updated on 01/Nov/20

$$Iaminclass11\left(highschool\right):)$$

Commented by prakash jain last updated on 02/Nov/20

$$\mathrm{You}\mathrm{are}\mathrm{doing}\mathrm{very}\mathrm{well}\mathrm{for}\mathrm{class}\mathrm{XI}.$$$$\mathrm{Are}\mathrm{you}\mathrm{aiming}\mathrm{for}\mathrm{IIT}?$$

Commented by Dwaipayan Shikari last updated on 02/Nov/20

$$Iwanttostudyphysicsinlater.$$$${IIT}^{\prime }saregoodbutiwanttostudyfreely$$

Commented by Dwaipayan Shikari last updated on 02/Nov/20

$$Butin2022ihavetotakethisexam:)$$

Commented by prakash jain last updated on 02/Nov/20

$$\mathrm{Are}\mathrm{u}\mathrm{interested}\mathrm{in}\mathrm{theoritical}\mathrm{physics}?$$

Commented by TANMAY PANACEA last updated on 02/Nov/20

$${17}^{1983}+{11}^{1983}-7^{1983}$$$$answeris=1$$$$explanation$$$$7^1=lastdigit7$$$$7^2=l.d9$$$$7^3=l.d3$$$$7^4=1$$$$7^5=7$$$$cycleof\left(7917\right)$$$$\frac{1983}{7}=283+\frac{2}{7}\to lastdigit=9$$$$lastdigitof{7}^{1983}\to =9$$$$lastdigitof{11}^{1983}is1$$$$so(9+1-9)\to lastdigit=1$$$$\mathit{p}\mathit{l}\mathit{s}\mathit{c}\mathit{h}\mathit{k}$$

Commented by TANMAY PANACEA last updated on 02/Nov/20

$${32}^{{32}^{32}}devidedby7$$$${(4\times 7+5)}^{{32}^{32}}$$$$=f\left(7\right)+5^{{32}^{32}}$$$$now{32}^{32}=evennumber$$$${32}^{32}$$$$2^1\to 2$$$$2^2\to 4$$$$2^3\to 8$$$$2^4\to 16$$$$2^5\to 32$$$$cycle\left(lastdigit2486\right)$$$${32}^{32}=2kisaevennumberwhoselastdigit6$$$$5^{2k}=lasttwodigitis25$$$$sowhendevidedby7\to lastdigitis4$$$$\mathit{p}\mathit{l}\mathit{s}\mathit{c}\mathit{h}\mathit{k}\mathit{i}\mathit{s}\mathit{m}\mathit{y}\mathit{t}\mathit{h}\mathit{i}\mathit{n}\mathit{k}\mathit{i}\mathit{n}\mathit{g}\mathit{c}\mathit{o}\mathit{r}\mathit{r}\mathit{e}\mathit{c}\mathit{t}$$$$$$$$$$$$$$

Commented by Dwaipayan Shikari last updated on 02/Nov/20

$$yes,Prakashsir!$$

Commented by Dwaipayan Shikari last updated on 03/Nov/20

$$(4\times 7+4)={32}^{{32}^{32}}$$$$typo$$

Commented by TANMAY PANACEA last updated on 03/Nov/20

$$yesyes$$

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