b and c 5 Prove by mathematical induction that for all positive integer values of n:(п + 1)! — 1b 2 x 1! + 5 × 2! + 10 × 3! + ·.. + (n² + 1)n! = n(n + 1)!1 x 1! + 2 × 2! + 3 × 3! + · ·. + n × n! = (n + 1)! - 11C2!23++ :.: +4!n1= 1(п + 1)!...3!(n + 1)!

(b) Case 1 : n=1 2×1!+5×2!+10×3!+...+(12+1)1! = 1(1+1)!2×1! = 1(2)!2 = 2 Therefore, the result is…

Answered: b 2x 1! + 5x 2! + .+ (nt + 1)п! %3D п(п…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Related questions

Question

b and c

5 Prove by mathematical induction that for all positive integer values of n:
(п + 1)! — 1
b 2 x 1! + 5 × 2! + 10 × 3! + ·.. + (n² + 1)n! = n(n + 1)!
1 x 1! + 2 × 2! + 3 × 3! + · ·. + n × n! = (n + 1)! - 1
1
C
2!
2
3
+
+ :.: +
4!
n
1
= 1
(п + 1)!
...
3!
(n + 1)!

Expert Solution

Step 1

(b) Case 1 : $n = 1$

$\begin{array}{rcl} 2 \times 1! + 5 \times 2! + 10 \times 3! + . . . + (1^{2} + 1) 1! & = & 1 (1 + 1)! \\ 2 \times 1! & = & 1 (2)! \\ 2 & = & 2 \end{array}$

Therefore, the result is true for $n = 1$

Case 2 : Assume result is true for $n = k$

Therefore $2 \times 1! + 5 \times 2! + 10 \times 3! + . . . + (k^{2} + 1) k! = k (k + 1)!$

Case 3 : Consider $n = k + 1$

Then

$\begin{array}{rcl} L H S & = & 2 \times 1! + 5 \times 2! + 10 \times 3! + . . . + (k^{2} + 1) k! + ({(k + 1)}^{2} + 1) (k + 1)! \\ = & k (k + 1)! + ({(k + 1)}^{2} + 1) (k + 1)! \\ = & (k + 1)! (k + {(k + 1)}^{2} + 1) \\ = & (k + 1)! (k + k^{2} + 2 k + 1 + 1) \\ = & (k + 1)! (k^{2} + 3 k + 2) \\ = & (k + 1)! (k + 2) (k + 1) \\ = & (k + 1) (k + 2)! \\ = & R H S \end{array}$