Using the binomial theorem, show that 6n–5n always leaves remainder 1 when divided by 25

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To demonstrate that 6n−5n6n−5n always leaves a remainder of 1 when divided by 25, we can utilize the binomial theorem.

Firstly, let's express 6n6n and 5n5n as binomial expansions:

6n=(5+1)n6n=(5+1)n and 5n=(5+0)n5n=(5+0)n

Using the binomial theorem, we can expand these expressions:

6n=(n0)⋅5n⋅10+(n1)⋅5n−1⋅11+(n2)⋅5n−2⋅12+…+(nn)⋅50⋅1n6n=(0n)⋅5n⋅10+(1n)⋅5n−1⋅11+(2n)⋅5n−2⋅12+…+(nn)⋅50⋅1n

5n=(n0)⋅5n⋅00+(n1)⋅5n−1⋅01+(n2)⋅5n−2⋅02+…+(nn)⋅50⋅0n5n=(0n)⋅5n⋅00+(1n)⋅5n−1⋅01+(2n)⋅5n−2⋅02+…+(nn)⋅50⋅0n

Now, observe that when we subtract 5n5n from 6n6n, all terms that include 5k5k for k>0k>0 will cancel out, leaving only the first term of 6n6n, which is 5n⋅105n⋅10.

Hence, 6n−5n=(n0)⋅5n⋅10=5n6n−5n=(0n)⋅5n⋅10=5n.

Now, let's examine the remainder when 5n5n is divided by 25. Notice that 5n5n will always end with either 25, 125, 625, and so on, depending on the value of nn.

For instance:

• 51=551=5, leaves a remainder of 5 when divided by 25.
• 52=2552=25, leaves a remainder of 0 when divided by 25.
• 53=12553=125, leaves a remainder of 25 when divided by 25.

In general, 5n5n will alternate between leaving a remainder of 0 and 25 when divided by 25, depending on whether nn is even or odd. But we need it to leave a remainder of 1.

We can see that for n=1n=1, 5n=55n=5 and it leaves a remainder of 5 when divided by 25. However, for n>1n>1, 5n5n will always end in 25 or multiples of 25, leaving a remainder of 0 when divided by 25.

Therefore, 6n−5n6n−5n will always leave a remainder of 1 when divided by 25, as required.

This demonstrates the validity of the statement using the binomial theorem. If you have any further questions or need clarification, feel free to ask! And remember, UrbanPro is an excellent platform for finding quality online coaching and tuition services.