120 (s+2)6 2s + 5 s(s – 3)2

Q: b) (-1)"3n² 4n²–1 n=1

A: We will use root test.

Q: s+2s ds s2+1

A: I have stated the used formulas in the solution.

Q: -5s – 36 16. (s + 2)(s² + 9) 3s2 + 5s + 3 18. s4 + s3

A: You asked only 16. Factoring in real field.

Q: 16. Σ 3k-1 k=1

A:

Q: 2-5s 5. (s-6)(s²+16)

A:

Q: 1OTO 6 -1 20 P = 0 -2 2

A:

Q: 8s - 16 (s² + s)(s² + 1),

A:

Q: 2 2 n(n-1)x", n²-n 37

A:

Q: 3s – 7 s- 2s + 3

A:

Q: 2 + 3s (s + 2)³ = (-21² + 3t) e-2i

A: Follow the procedure given below.

Q: s2-2s -n: (s-+4)(s-+4s+5)

A: Given: s2-2ss2+4s2+4s+5

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

A:

Q: -2 メ S-1 2メ-1 <メくS こa 2メ 0<メン メ メ フ」

A: we have to draw the graph

Q: 2s-3 1. s2-4s+8

A:

Q: 2s² 2s - bs +5 (s-1)(s-2) (s- 25.4 (s+)(s-2)(s-3)

A:

Q: 3s-27 6s-35 H 9s-80 4s-42 K %3= Submit

A:

Q: 2n - 1 Σ n2 +n + 4 n=1

A:

Q: s2-4 ds 2-s6

A: We are given an indefinite integral which cannot be reduced into a proper functional form. Hence we…

Q: 3s2 + 5s + 1 (s)= (s2 + 15s + 34) %3D

A:

Q: S(z^3){[9+(z^2)]^(1/3)}dz

A:

Q: 38. 32 (-1)k k=0 (2k + 1)³

A:

Q: a1 = -31, d = -7, n = 5 O-231 0-230 O1-225 -240

A:

Q: ´5s²–15s-11) -(s+1)(s–2)³. 34-2s 2. i) Find ii) Evaluate (15(s-5)(s+3).

A: The solution is given as

Q: 8. P(-0.9 <Z < 0.19)

A: (8) z be a standard normal curve.z~N(μ,σ)where   ,  mean(μ)=0               standard…

Q: 123 A= 4 5 6 7 8 9 -1 7 3 B= 4 -2 5 -3 9 4 13 2 C= 6. 0 10 3 6. DD 2 L-7 8 4 F- 8. %3D

A:

Q: 9-12s-4s2 C. (s2+4)2 +12s-2s2 (s2+4)2 9+12s-4s? D. (s2+4)3 -12s-2s? (s2+4)3

A:

Q: (n² + 6n – 4)(2n – 4)

A: The polynomial is given by n2+6n-4(2n-4)

Q: n(n + 1)(2n + 1) 6 k=0

A: In this question, we prove the sum of squares of natural numbers.

Q: 2(n!)2 (2n)! n=1

A:

Q: 2. 1. 2 8. 2. -2 そ %3D 00 11 元N ゴ

A:

Q: 5s' -7s+17 b- 1". (s-1)(s' +4)

A: First of all we will do partial fraction, then we will apply kappas inverse

Q: 2) (n²-n-29) (n-6)

A: Please refer to the image below

Q: n! 00 is 2.5.8....(3n+2) Zn=1 а. 3 b. 5 с. d. 3.

A: For the following problem,

Q: 4β) 5 482/25 28 -10-t(n-1)

A: Random variable can be α=4,β Mean and variance can be, E(X)=αβ=4βVar(X)=αβ2=4β2 Size of sample is…

Q: Σ (-3)" 3.5.7... (2n + 1) n=1 8.

A:

Q: | grk(1 – 2)n=k/(") z*(1 – 2)n-k | 2*(1 – : 1/2 1 2n/2 how did we get step 3 from step 2 ? explain…

A: First consider the second term of step 2 , which is ∑k=0nnkxk1-xn-k12 This term is in the form of…

Q: 2(-1)n-1_6" n=1 5n+1

A: To find: ∑n=1∞-1n-16n5n+1

Q: 2n 2 3n2 +4 n=1

A:

Q: n Σ k + 2n + n2 | n k=1

A: use the sum of first natural number  to evaluate the sum

Q: 2s – 4 (s² + s)(s² + 1) 2

A:

Q: a)(6n2– 3n + 9) - (n² + 3n – 5) b) (15a – 3w + 7) – 2(2a – 9) + (3w – 15a)

A: a) Given:         6n2-3n+9-n2+3n-5       =6n2-3n+9-n2-3n+5 combine like terms…

Q: Σ (-3)" 3.5.7..· (2n + 1) n=1

A:

Q: Σ 2n+3 + n2 +3n n-1 3.

A: Given series ∑n=1∞1n2+3n+2n+33n

Q: S (2n)!n² (2n + 2)!

A: We have to find  limn→∞(2n)! n2(2n+2)!

Q: (ii)""+2n-1 n' +3

A: We have to test the series is convergent or divergent

Q: Q.5) (a) Evaluate L-! 2+6s+9 (3-1)(s-2)(s–4

A: First use partial fraction then take inverse laplace transform

Q: s2 3- L-1 (s-2)2.

A: Take a partial fraction

Q: (b) (-1)"n (n+1)(n+2)

A:

Q: 2) . (4а — 2)2n+1 n! n=0

A: Given series is ∑n=0∞4x-22n+1n!

Q: -4:General Form 2 3.

A: