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1h
answered $\lim_{n\to\infty} \dfrac{a^n}{n!} = 0$
6h
answered $ (3+\sqrt{5})^n+(3-\sqrt{5})^n\equiv\; 0 \; [2^n] $
7h
answered Computing the limit of a summation of sequence
17h
answered How to calculate this limit without De L'Hopital?
17h
comment Find monic quartic polynomial f(x) with rational coefficients whose roots include…(Algebra)
For polynomial equations with rational coefficients, the irrational roots occur in conjugate pair See docs.google.com/document/d/…
18h
revised Find a closed form for the equations $1^3 = 1$, $2^3 = 3 + 5$, $3^3 = 7 + 9 + 11$
added 91 characters in body
18h
revised Find a closed form for the equations $1^3 = 1$, $2^3 = 3 + 5$, $3^3 = 7 + 9 + 11$
added 91 characters in body
19h
comment Find a closed form for the equations $1^3 = 1$, $2^3 = 3 + 5$, $3^3 = 7 + 9 + 11$
@nbro, The question should be how, not why:)How can you find the sum unless you derive the first term?
19h
comment Prove that one integer among $m$ consecutive integers is divisible by $m$
@PranavA.R, Please find the edited version
19h
revised Prove that one integer among $m$ consecutive integers is divisible by $m$
added 415 characters in body
19h
answered Find a closed form for the equations $1^3 = 1$, $2^3 = 3 + 5$, $3^3 = 7 + 9 + 11$
19h
answered Proof $\displaystyle \binom{p-1}{k}\equiv (-1)^k \mod{p}$
1d
comment Prove that one integer among $m$ consecutive integers is divisible by $m$
@PranavA.R, Hope you don't have any confusion as you accepted the answer
1d
answered Does a system om congruence equations have solutions?
1d
answered Solving $\cos^6(2x)+\sin^6(2x)=\frac58$
1d
revised Solving $\cos^6(2x)+\sin^6(2x)=\frac58$
deleted 4 characters in body
1d
answered Solving $\cos^6(2x)+\sin^6(2x)=\frac58$
1d
answered Calculate sum of angles if you know their tan value
1d
comment Prove that one integer among $m$ consecutive integers is divisible by $m$
@ZubinMukerjee, Let none of them is divisible by $m$. Then there can be $m-1$ distinct remainders $1\cdots m-1$ by $m$ $b_r$s
1d
comment Prove that one integer among $m$ consecutive integers is divisible by $m$
@ZubinMukerjee, Updated. Thanks for your observation