ised. (a) Ž 3n2+1 5n2+3 n=1

Q: )"-( 18 6. lim V n4n +(4n)" 2+ n2 4 4+ n2

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Q: Determine whether or not the infinite series below converges.

A: Since you have asked multiple questions, we will solve the first question for you. If you want any…

Q: )- ニ ZS+ト 32 |ニ2+

A: To solve the equation: x+y+z=12x-4y-3z=9-3x+2y+5z=-5

Q: ( a ) Ση-1 (b) En=2 n7-1 15n-2 (c) En=1 11n+3 1 (d) Σn1 n14 (-1)" (e) En=1 7n+4

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Q: n2 – n 1 2n² + n الم نابعة زختبر تقارب -

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Q: The solution of e² = - 11 has the form x + ily + 2nn) where n is an integer. What is the value of x?…

A: The given equation is                                                ez=-11

Q: (a) an = }(n² + n). (b) an = 2an-1 – an-2 with ao = 0 and aj = 1. (c) an = nan-1 with ao = 1. %3D

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Q: 7. Find k > 0 and m E N so that 6n3 + 17n m.

A: The solution of given problem as follows :

Q: 26 E(- 1)"2n - %3D n=0

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Q: 31. x2 + 2x - 12 = 0 34. 2x + x - 3 = 0 37 3x? - 6x – 9 = 0 40. x2 - 6x + 18 = 0 43. 2x? - 7x = -12…

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Q: -6-2cos 0 = -4 + 2cos 0 %3D T 7n B) 3' 3 4. 2n 47 n 2n C) 4 D) { d, 3. 3 4.

A: Given expression: -6-2cosθ=-4+2cosθ Simplify the above equation…

Q: n-1 For all n > 2, 4+ 3n4" – 4n+1 - j=1 9.

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Q: a) 5n + 1 3n2 2" (b) Σ 1 ( c) Σ n=2 nº - 7 - 1 n=1 n=1 3n

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Q: The solution of e² = -2 for n = 0,±1,±2,. is %3D ... Select one: O a. In2 + in(1+n) O b. In2 +…

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Q: (-1)**'n! 1-3-5..(2n+1) n+1

A: The objective is to find whether the series is convergent or divergent.

Q: (-1)n+13n (b) n5n n=1

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Q: 2*-2 2. E 23- (n² +4) 6n² 3. 2 n=0 N° - 5n +3

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Q: (-1)"(n + 2) 2n+1(n + 1) 4. Compute the exact value of n=0

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Q: 2n [3] (b) E for n 21 2k k=0

A: Answer

Q: D. Show that if an 3n3 - 17n2 + 2n - 6, then a, = O(n³).

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Q: 3. The value of c is given by Vn А. /n+1

A: This is a limit problem

Q: (-1)"n²(x+3)" 2. 5n n=1

A: Radius of convergence

Q: 2. Find the value of 2n=1 Sn=15 2n + 3.

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Q: 1+n+ Vn n=0 n+ Vn4 + n² +1 O A. 32 O B. 2-8 O C. - 16

A: For checking, whether the series is convergent or divergent, we can directly check the highest power…

Q: -ges 3n +n 9n b) L7-2)n+1n n! n=1 (-10)" d) 242n+1(n+1) n2 (2n – 1)! n=1 0=

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Q: (-1)"(n+ 2) 2n+1(n + 1) 4. Compute the exact value of n=0

A: Given series is  ∑n=0∞-1nn+22n+1n+1

Q: Derive a formula for fx* In x dxwhere a.) n+ -1 b.) n = -1

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Q: - 31 0 -6 4 T3 1 0 07 - Let A 0 -1 -7 B 4 8 -4 0 - – . 4 1 -4 3 - |AB| = |BC| = II

A: Hi, thank you for the question. Since you have posted multiple questions as per our policy we will…

Q: 5) h(n) =-2n² + 4; Find h(4)

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Q: 2n zn こ (n) n:0

A: note : as per our guidelines we are supposed to answer only one question. Kindly repost other…

Q: Calculate the convolution: x * 2 * e2* . x2 e2a O a. 4 4 8 e2a Ob. 4 4 8 - 4 2 4 O d. 1 x2 e2æ 4 2 4…

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Q: 4n +1 (a) 2n2 +3 n=0 5n2 – n+1

A: Since multiple questions are asked, I am answering first question

Q: (r+ simo - coso)dr+r Csino tcoo + co-B

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Q: Identify the two series that are the same.

A: We have to find which series are same. We have given series…

Q: 2. Show that coo(N) co(N) § c(N) Ę e^(N).

A: To prove that c00N⊂c0N⊂cN⊂l∞N. By definition, c00=x∈lp|all but finitely many xj's are zeroc=x∈l∞|xj…

Q: 2" (a) 3 3n+1 (c) E(-1)"; 4n n=1 (b) 3+5n² (4) E n+ n2 n=1 n=1

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Q: (-1)" (n+ 2) 2n+1(n + 1) Compute the exact value of

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Q: 3 n' (iv) En=1 2n³+5

A: (iv). Series  ∑n=1∞n32n3+5 diverges by divergence test for series.

Q: 4h+7 3) Ž 00 2) 2 nñ 二 n3+5n²-17 n en n n=1 n =2

A: NOTE : As per Bartleby guidelines for more than one question asked only one is to be answered please…

Q: ¸1(1 + r)n−¹ − 1. (n-1-1₁ r Sn (1+r)"So - P[- =

A: Sn = 1+rnS0 - P1+rn-1-1rAim: To find expression for 'n'.…

Q: 7|32n+1 + 2n+2, Vn E Z.

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Q: (d) (e) n=1 n=1 3n²+ 1 2n² +3 2¹ (n!)² n2(2n!)

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Q: 2+4+... +n lim 7n2 + yn

A: limn→∞2+4+....+n7n2+n

Q: Find the possible values of n. (n2 )" – 2n" + 1 = 0 (n +0,1)

A: We have  n2n-2nn+1=0 and n≠0,1 We can write n2n as n2n=n2n=nn2 So, nn2-2nn+1=0 Let nn=x x2-2x+1=0

Q: (b) Z/n = (q) if and only if kq 0 for all 1 < k <n.

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Q: (-1)".n -1)".n 5n+2

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Q: 1 Vn > 0, 2(аn+1 — аn) %- 2n

A: Given an=1-12n so an+1=1-12n+1.

Q: 1. Σ n! n=0 (-1)"x²n+1 2. ) 2n + 1 n=0 (-1)"x2n 3.) (2n)! n=0 x2n+1 4. (-1)"

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Q: (2п)! Zn=1 2n.n!

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