Solution:
To prove: 3^n + 4^n 5^n for all n 2
Explanation:
Proving using Mathematical Induction:
Step 1:
=>Checking for base condition n = 2
Put n = 2
=>LHS = 3^2 + 4^2
=>LHS = 9 + 16
=>LHS = 25
=>RHS = 5^2
=>RHS = 25
=>Hence LHS = RHS for base condition so 3^n + 4^n 5^n is true for base condition.
Step 2:
=>Assuming the expression 3^n + 4^n 5^n is true for n = k
=>3^k + 4^k 5^k....(1)
Step 3:
=>Checking the expression for n = k+1
=>LHS = 3^(k+1) + 4^(k+1)
=>LHS = 3*3^k + 4*4^k
=>LHS = (3^k + 4^k) + 2*3^k + 3*4^k
=>Taking equal(=) sign from equation (1) so we can write 3^k + 4^k = 5^k
=>LHS = 5^k + 2*3^k + 3*4^k
=>RHS = 5^(k+1)
=>RHS = 5*5^k
=>RHS = 5^k + 4*5^k
=>We know that expression 4*5^k > 2*3^k + 3*4^k
=>Hence we can write 5^k + 2*3^k + 3*4^k 5^k + 4*5^k
=>Hence the expression is true for n = k+1
=>Hence on the basis of above statements we have proved our expression using Mathematical Induction.
I have explained each and every part with the help of statements attached to it.
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