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Th is question is on page 78 of the C2 book if more information is needed.

Q.10 (question number is in the book), part a.

In the binomial expansion of (2k+x)^n, where k is a constant and n is a positive integer, the coefficient x^2 is equal to the coefficient of x^3.

Prove that n= 6k+2

-----------------------------------------------------------

Q.13 part b

Given that (2+x)^5 +(2-x)^5 = (indentical) A + Bx^2 +Cx^4

I found that A=64 B=160 C=20

Now using the the substitution of y= x^2 and the answers to part a, solve

(2+x)^5 + (2-x)^5=349

i dont understand the part about subtituting y= x^2

Thanks.

Q.10 (question number is in the book), part a.

In the binomial expansion of (2k+x)^n, where k is a constant and n is a positive integer, the coefficient x^2 is equal to the coefficient of x^3.

Prove that n= 6k+2

-----------------------------------------------------------

Q.13 part b

Given that (2+x)^5 +(2-x)^5 = (indentical) A + Bx^2 +Cx^4

I found that A=64 B=160 C=20

Now using the the substitution of y= x^2 and the answers to part a, solve

(2+x)^5 + (2-x)^5=349

i dont understand the part about subtituting y= x^2

Thanks.

You sure the first question is correct? I keep getting n = 2k + 2.

ScipioAfricanus

Q.13 part b

Given that (2+x)^5 +(2-x)^5 = (indentical) A + Bx^2 +Cx^4

I found that A=64 B=160 C=20

Now using the the substitution of y= x^2 and the answers to part a, solve

(2+x)^5 + (2-x)^5=349

i dont understand the part about subtituting y= x^2

Thanks.

Given that (2+x)^5 +(2-x)^5 = (indentical) A + Bx^2 +Cx^4

I found that A=64 B=160 C=20

Now using the the substitution of y= x^2 and the answers to part a, solve

(2+x)^5 + (2-x)^5=349

i dont understand the part about subtituting y= x^2

Thanks.

Assuming your values of A, B and C are correct:

64 + 160y + 20y² = 349

20y² + 160y - 285 = 0

4y² + 32y - 57 = 0

(2y + 19)(2y - 3) = 0

=> y = -19/2 or 3/2.

But, y = x², so y > 0 if x E R.

Hence, y = 3/2.

Substituting:

x² = 3/2

x = ±(3/2)^0.5

Hope this helps,

~~Simba

Thanks guys.

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