Given the recurence relation x = 10r, - 25x-1 and xg = 1,x = 10. Weneed to prove that x, = 5"(1 + n) for all n 2 1. To do so, the inductive stepshould beXn = 5"(1+n) and showthat xm+1 = 5"+1(2+n).By first principle of mathematicalO induction, we need to suppose theaboveXg+1 = 10xx – 25xg -1 for k < nand show that x+1 = 10xn – 25xn-1 ·By Strong induction, we need tosuppose the aboveNone of theseXg = 5*(1+ k) for k < nand show that x, = 5"(1+n).By Strong induction, we need tosuppose the aboveXn = 5"(1+) and showthat xn+1 = 5"+1(2 +n).By Strong induction, we need tosuppose the above

Given,xn+1=10xn-25xn-1 and x0=1, x1=10. To prove: xn=5nn+1; n≥1. By first principlw of induction…

Geven the recmence nelation = 10r,-25 al= L, = 10. We meed to prove that , =5"(1+R) for all21. To do so, the inductive step shold be Xn = 5"(1+ n) and show that x+1 = 5"+'(2 +n). By first principle of mathematical induction, we need to suppose the above X*+1 = 10x - 25x-1 for k

Geven the recmence nelation = 10r,-25 al= L, = 10. We meed to prove that , =5"(1+R) for all21. To do so, the inductive step shold be Xn = 5"(1+ n) and show that x+1 = 5"+'(2 +n). By first principle of mathematical induction, we need to suppose the above X*+1 = 10x - 25x-1 for k

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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Numerical Integration

In a mathematical investigation, numerical integration comprises a wide group of calculations for computing the mathematical estimation of definite integral, and likewise, the term is moreover in some cases used to depict the numerical solution of differential equations. The most important aspect of this theory is error analysis.

Question

Given the recurence relation x = 10r, - 25x-1 and xg = 1,x = 10. We
need to prove that x, = 5"(1 + n) for all n 2 1. To do so, the inductive step
should be
Xn = 5"(1+n) and show
that xm+1 = 5"+1(2+n).
By first principle of mathematical
O induction, we need to suppose the
above
Xg+1 = 10xx – 25xg -1 for k < n
and show that x+1 = 10xn – 25xn-1 ·
By Strong induction, we need to
suppose the above
None of these
Xg = 5*(1+ k) for k < n
and show that x, = 5"(1+n).
By Strong induction, we need to
suppose the above
Xn = 5"(1+) and show
that xn+1 = 5"+1(2 +n).
By Strong induction, we need to
suppose the above

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