As an experienced tutor registered on UrbanPro, I'm well-equipped to guide you through this topic. UrbanPro is indeed an excellent platform for online coaching and tuition. Now, let's delve into expanding the expression (2x - 3)^6 using the binomial theorem.

The binomial theorem is a powerful tool that allows us to expand expressions of the form (a + b)^n, where 'a' and 'b' are any real numbers, and 'n' is a positive integer.

In this case, our expression is (2x - 3)^6. According to the binomial theorem, we can expand this expression as follows:

(2x - 3)^6 = C(6, 0)(2x)^6(-3)^0 + C(6, 1)(2x)^5(-3)^1 + C(6, 2)(2x)^4(-3)^2 + ... + C(6, 6)(2x)^0(-3)^6

Here, C(n, k) represents the binomial coefficient, also known as "n choose k", which is calculated as n! / (k! * (n - k)!), where '!' denotes factorial.

Let's simplify each term:

1. C(6, 0)(2x)^6(-3)^0 = 1 * (2x)^6 * 1 = 64x^6

2. C(6, 1)(2x)^5(-3)^1 = 6 * (2x)^5 * (-3) = -648x^5

3. C(6, 2)(2x)^4(-3)^2 = 15 * (2x)^4 * 9 = 4320x^4

4. Continuing this pattern, we find all terms until:

5. C(6, 6)(2x)^0(-3)^6 = 1 * 1 * (-729) = -729

Now, let's collect all the terms:

64x^6 - 648x^5 + 4320x^4 - ... + (-729)

And there you have it! The expansion of the expression (2x - 3)^6 using the binomial theorem. If you have any further questions or need clarification on any step, feel free to ask!

As an experienced tutor registered on UrbanPro, I'd be delighted to help you with this question.

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.

As an experienced tutor registered on UrbanPro, I can confidently say that UrbanPro is one of the best platforms for online coaching and tuition services. Now, let's dive into expanding the expression (a + 1/b)^11 using the Binomial Theorem.

The Binomial Theorem states that for any positive integer n, the expansion of (x + y)^n is given by:

(x + y)^n = Σ [nCr * x^(n-r) * y^r]

Where Σ denotes summation from r = 0 to n, and nCr represents the binomial coefficient, also known as "n choose r."

In this case, we have (a + 1/b)^11. Applying the Binomial Theorem:

(a + 1/b)^11 = Σ [11Cr * a^(11-r) * (1/b)^r]

Now, let's compute the coefficients and simplify the expression:

= 11C0 * a^11 * (1/b)^0 + 11C1 * a^10 * (1/b)^1 + ... + 11C11 * a^0 * (1/b)^11

Now, let's calculate each term:

= a^11 + 11 * a^10 * (1/b) + 55 * a^9 * (1/b)^2 + ... + (1/b)^11

This expansion continues until the last term. Let's simplify further:

= a^11 + 11a^10/b + 55a^9/b^2 + ... + 1/b^11

And there you have it, the expansion of (a + 1/b)^11 using the Binomial Theorem. If you need further clarification or assistance, feel free to ask! And remember, UrbanPro is here to provide excellent tutoring services for your academic needs.