What is the solution to the Binomial Theorem problem highlighted in red?

[Miike012]

I highlighted the portion in red in the paint document that I'm not understanding.

How can we see by inspection that the product is equal to the series 2?

 

[pasmith]

I highlighted the portion in red in the paint document that I'm not understanding.

How can we see by inspection that the product is equal to the series 2?

The term independent of x in the product is equal to the series (2); this is because the term independent of x is a sum of terms of the form $kc_k x^k \times c_k/x^k = kc_k^2$.

 

[Miike012]

The term independent of x in the product is equal to the series (2); this is because the term independent of x is a sum of terms of the form $kc_k x^k \times c_k/x^k = kc_k^2$.

The second line in the paint document is equal to n/xⁿ(1 + (2n-1)x + (2n-1)(2n-1)/2!x² + ...) by using the binomial theorem

If we look at the coefficient of the second term it is equal to n(2n-1).

If we compare the coeff. n(2n -1) with the coeff. of the second term of series (2) which is
2c₂² = 2(n-1)(n-2)/2! = n² - 3n + 2 they are not equal.

Hence n(2n-1) =/= n² - 3n + 2

 

[Miike012]

Never mind I see that u said its the sum of the terms... I got it now