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Question Number 159675 Answers: 1 Comments: 2

Question Number 159560 Answers: 1 Comments: 0

Resolve 1. u_(n+2) −2u_(n+1) +4u_n =3^n with u_o =1, u_1 =−2 2. u_n =u_(n−1) −u_(n−2) +2sin (((nΠ)/3)) with u_o =1, u_1 =2

$R e s o l v e$ $1 . u_{n + 2} - 2 u_{n + 1} + 4 u_{n} = 3^{n}$ $w i t h u_{o} = 1, u_{1} = - 2$ $2 . u_{n} = u_{n - 1} - u_{n - 2} + 2 \sin (\frac{n Π}{3})$ $w i t h u_{o} = 1, u_{1} = 2$

Question Number 159475 Answers: 0 Comments: 0

Question Number 159425 Answers: 0 Comments: 0

Question Number 159403 Answers: 0 Comments: 1

U_(n+1) =(1/2)(u_n +(a/u_n )) with u_1 >0, a>0 Prove that (u_(n+1) /u_n )≤1

$U_{n + 1} = \frac{1}{2} (u_{n} + \frac{a}{u_{n}}) w i t h u_{1} > 0, a > 0$ $P r o v e t h a t \frac{u_{n + 1}}{u_{n}} ⩽ 1$

Question Number 159309 Answers: 1 Comments: 0

Resolve I_n =∫_(−1) ^1 (1−x^2 )^n dx

$R e s o l v e I_{n} = \int_{- 1}^{1} {(1 - x^{2})}^{n} d x$

Question Number 159171 Answers: 1 Comments: 0

Prove by absurd that log 2 is the number irrational

$P r o v e b y a b s u r d t h a t \log 2 i s t h e$ $n u m b e r i r r a t i o n a l$

Question Number 159123 Answers: 2 Comments: 0

Question Number 159078 Answers: 0 Comments: 0

1) Prove by recurrence that for n≥28, n!≥11^n 2) On subtract the limit of the suite (((n!)/(10^n ))) when n tended at +∞

$1) P r o v e b y r e c u r r e n c e t h a t$ $f o r n ⩾ 28, n! ⩾ 11^{n}$ $2) O n s u b t r a c t t h e l i m i t o f t h e$ $s u i t e (\frac{n!}{10^{n}}) w h e n n t e n d e d a t + \infty$

Question Number 159016 Answers: 0 Comments: 0

Question Number 158984 Answers: 0 Comments: 0

1. Prove by recurrence that so n ∈ N and θ ∈ R (cos (nθ)+isin (nθ)=cos (nθ)+isin (nθ) 2. Prove that U_(n+1) =(1/5)(U_n ^2 +6) and U_1 =(5/2), is decrease

$1 . P r o v e b y r e c u r r e n c e t h a t$ $s o n \in N a n d θ \in R$ $(\cos (n θ) + i \sin (n θ) = \cos (n θ) + i \sin (n θ)$ $2 . P r o v e t h a t U_{n + 1} = \frac{1}{5} (U_{n}^{2} + 6) a n d$ $U_{1} = \frac{5}{2}, i s d e c r e a s e$

Question Number 158945 Answers: 2 Comments: 0

1) Prove by absurd that ((ln 2)/(ln 3)) is irrational 2) Prove by absurd that (√2)+(√(6 ))≤(√(15))

$1) P r o v e b y a b s u r d t h a t$ $\frac{\ln 2}{\ln 3} i s i r r a t i o n a l$ $2) P r o v e b y a b s u r d t h a t$ $\sqrt{2} + \sqrt{6} ⩽ \sqrt{15}$

Question Number 158858 Answers: 0 Comments: 0

I_n =∫_(−1) ^1 (1−x^2 )^n cos ((a/(2b))x)dx to integrating by piece for n≥2 proven (a^2 /(4b^2 ))I_(n ) =2n(2n−1)I_(n−1) −4(n−1)I_(n−2) proven by rearring that ((a/(2b)))^(2n+1) I_n =n![p((q/(2b)))sin ((a/(2b)))+Q((a/(2b)))cos ((a/(2b)))]

$I_{n} = \int_{- 1}^{1} {(1 - x^{2})}^{n} \cos (\frac{a}{2 b} x) d x$ $t o i n t e g r a t i n g b y p i e c e f o r n ⩾ 2$ $p r o v e n$ $\frac{a^{2}}{4 b^{2}} I_{n} = 2 n (2 n - 1) I_{n - 1} - 4 (n - 1) I_{n - 2}$ $p r o v e n b y r e a r r i n g t h a t$ ${(\frac{a}{2 b})}^{2 n + 1} I_{n} = n! [p (\frac{q}{2 b}) \sin (\frac{a}{2 b}) + Q (\frac{a}{2 b}) \cos (\frac{a}{2 b})]$