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Bingel [31]
3 years ago
12

Use mathematical induction to prove that for each integer n ≥ 4, 5^n ≥ 2 2^n+1 + 100

Mathematics
1 answer:
anastassius [24]3 years ago
4 0

The given Statement which we have to prove using mathematical induction is

   5^n\geq 2*2^{n+1}+100

for , n≥4.

⇒For, n=4

LHS

=5^4\\\\5*5*5*5\\\\=625\\\\\text{RHS}=2.2^{4+1}+100\\\\=64+100\\\\=164

 LHS >RHS

Hence this statement is true for, n=4.

⇒Suppose this statement is true for, n=k.

 5^k\geq 2*2^{k+1}+100

                      -------------------------------------------(1)

Now, we will prove that , this statement is true for, n=k+1.

5^{k+1}\geq 2*2^{k+1+1}+100\\\\5^{k+1}\geq 2^{k+3}+100

LHS

5^{k+1}=5^k*5\\\\5^k*5\geq 5 \times(2*2^{k+1}+100)----\text{Using 1}\\\\5^k*5\geq (3+2) \times(2*2^{k+1}+100)\\\\ 5^k*5\geq 3\times (2^{k+2}+100)+2 \times(2*2^{k+1}+100)\\\\5^k*5\geq 3\times(2^{k+2}+100)+(2^{k+3}+200)\\\\5^{k+1}\geq (2^{k+3}+100)+3\times2^{k+2}+400\\\\5^{k+1}\geq (2^{k+3}+100)+\text{Any number}\\\\5^{k+1}\geq (2^{k+3}+100)

Hence this Statement is true for , n=k+1, whenever it is true for, n=k.

Hence Proved.

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