Yes!😅

Fully agree.

the hardest thing is to remember this when you don't use it regularly

Hehehe 🎉

"best use" academically speaking. This teacher is Awesome. But, There are other Great things here on KZfaq.

Thank you! 😊

if u found it easy then ur prolly dumb

Happy to help!

​@@mrhtutoringsir I follow you but some things i don't understood

I didn’t learn it until college

I was hoping to see a cat...

Never knew this. Thank you for the topic

Wow, thanks. I appreciate your comment!

It's true

When your students who does not enjoy maths is inspired by your teaching you know you're a great teacher! Thank you so much!

I will try! Thank you

Thanks for subscribing!

Thank you for the additional information

Thank for the awesome comment. I appreciate it

Wow, thank you!

Glad it helped.

You're most welcome

No problem!

🙏

Fantastic!

You're very welcome!

Happy to see someone who appreciates math.

Thank you so much 😀

Welcome!

Thank you~

Thank you too!

Happy to help

Most welcome

It's my pleasure

Thanks for the comments! I appreciate them.

Music to my ears!

Yes~

This is way more interesting and easy to remember.

Agree

Yes. You just let beta = -b. Expand (a + beta)^5, exactly as you would with this method. Assuming a is positive: Every odd exponent on beta, will materialize as a negative term when translated back to b. Every even exponent on beta, will materialize as a positive term when translated back to b.

Good luck on your final!

I just thank you very much

I take it you are talking about convergence and divergence, of an infinite series, as there are other meanings of those words. A sequence (or progression) is a list of numbers, a series is a summation of a sequence. Usually k represents the index of the individual term, and n represents the total number of terms. There is usually a function of k that determines the sequence value, although a sequence can also be recursively defined like the Fibonacci sequence. Some series add up to a finite value, when you have an infinite number of terms, which is called a series that converges, or convergent. Other series either add up to infinity, or continuously flip/flop between the same two numbers and never get closer to zero or to infinity, which is called a divergent series. Some series include an alternator term, of (-1)^k. If an alternating series only converges when it is alternating, it's called conditionally convergent, for instance the harmonic series of 1/k. If a series converges, regardless of whether it alternates, it's called absolutely convergent. The substance of Calculus's work with series, is to analytically determine whether or not it converges, and if so, what value it converges to. There are various tests for convergence, based on properties of the series definition, that may use limits, powers, roots, and integrals, to test which kind of convergence a series has.

You can put a letter in that number's place. For your example, let t = 2. Thus: (d + t)^8 = d^8 + 8*d^7*t + 28*d^6*t^2 + 56*d^5*t^3 + 70*d^4*t^4 + 56*d^3*t^5 + 28*d^2*t^6 + 8*d*t^7 + t^8 8th row of Pascal's triangle for reference: 1, 8, 28, 56, 70, 56, 28, 8, 1 Now evaluate each t-power at t=2, and have it join the coefficient. Put the d-power aside for later. 8*t = 16 28*t^2 = 28*4 = 112 56*t^3 = 56*8 = 448 70*t^4 = 70*16 = 1120 56*t^5 = 56*32 = 1792 28*t^6 = 28*64 = 1792 8*t^7 = 8*128 = 1024 t^8 = 256 Thus, the result is: d^8 + 16*d^7 + 112*d^6 + 448*d^5 + 1120*d^4 + 1792*d^3 + 1792*d^2 + 1024*d + 256

Glad you found it easy.

Yes, I prefer the binomial theorem myself

atleast I got to know about it in school yesterday and now here 😂😂😂

Here's an example with a negative: Suppose it starts as (u - v)^3, where u and v are both positive numbers. Let w = -v. Now we've reduced it to (u + w)^3. Expand it out accordingly, with u's power descending from 3 to 0, and w's power ascending from 0 to 3. Coefficients follow 3rd row of Pascal's triangle, which is 1, 3, 3, 1. u^3 + 3*u^2*w + 3*u*w^2 + w^3 Since squared negatives are positive, this means all terms with w to an even power, will equal the same term with v replacing w. The opposite is true for cubed negatives. All terms with w to an odd power, will have a negative coefficient, and otherwise equal the same term. This gives us the result: u^3 - 3*u^2*v + 3*u*v^2 - v^3