Using Induction: Let n E N. Prove that if f(x) = x^n, then f^(n)(x) = n!. In other words, prove the
Formula (d/dx)^n x^n = n! for all n E N.
Solution
put n=1
(d/dx)^1*x^1 =1!
dx/dx=1
1=1
It is true for n=1
let n=k assume that it works
thus,(d/dx)^k (x)^k =k!
suppose n=k+1
(d/dx)(k+1) (x)^(k+1)
=>d/dx)^k d/dx (x)^(k+1)
=(d/dx)^k *(k+1) x^(k+1-1)
=>(k+1)(d/dx)^k *x^k
but (d/dx)^k *x^k =k!
substitue
(k+1)k!
(k+1)!
thus it is true for n=k+1
Therefore by principle of mathematical induction it is true for all x.
Using Induction prove = nn + 1 for n 1SolutionInduction basi.pdf
1. Using Induction prove = n/n + 1 for n 1
Solution
Induction basis:
n = 1 :
1/(i(i+1)) = 1/(1*2) = 1/2 = 1/(1+1)
Inductive step:
Assume that the statement holds for n = m:
m 1/(i(i+1)) = m/(m+1)
for n = m+1, we have:
m+1 1/(i(i+1)) = m 1/(i(i+1)) + 1/((m+1)(m+2)) = m/(m+1) + 1/((m+1)(m+2)) =
1/(m+1) (m + 1/(m+2)) = (m(m+2)+1)/((m+1)(m+2)) = (m^2 + 2m + 1)/((m+1)(m+2)) =
(m+1)^2/((m+1)(m+2)) = (m+1)/(m+2) = (m+1)/(m+1+1)
so the statement also holds for n = m+1.
Thus by mathematical induction it is true for all values of n.
