in the binomial expansion of (1+x/k)^n, where k is a constant and n is a positive integer, the coefficients of x and x^2 are equal.

(a) show that 2k=n-1
(b) deduce the value of k. Hence find the 1st 3 terms in the expansion in ascending powers of x.

(1+x/k)^n = 1^n + n(x/k) + n(n-1)/2 (x/k)^2

n/k = n(n-1)/2k^2
2kn/2k^2 = n(n-1)/2k^2
2kn = n(n-1)
2k = n-1

we know that n is odd, since n=2k+1
if k=1,
(1+x)^3 = 1+3x+3x^2+...

if k=2, we have
(1+x/2)^5 = 1+5(x/2) + 5*4/2*(x/2)^2+...
= 1 + 5/2 x + 5/2 x^2 + ...

For any integer value of k>0, n=2k+1
will work.

So, (b) is really a bogus question

(a) To find the coefficients of x and x^2 in the binomial expansion (1 + x/k)^n, we can use the formula for the coefficients given by Pascal's triangle. The coefficient of x is given by C(n, 1) = n, and the coefficient of x^2 is given by C(n, 2) = n(n - 1)/2.

Setting these two coefficients equal to each other, we have:

n = n(n - 1)/2

Multiply both sides by 2 to eliminate the fraction:

2n = n(n - 1)

Distribute n on the right side:

2n = n^2 - n

Rearranging this equation brings us to:

n^2 - 3n = 0

Factor out an n:

n(n - 3) = 0

From here, we have two possible solutions: n = 0 or n - 3 = 0. However, since n is defined as a positive integer, the only possible solution is n = 3.

Substituting n = 3 into the equation in the beginning:

2k = 3 - 1

2k = 2

Divide both sides by 2:

k = 1

Therefore, we have found that k = 1.

(b) Now that we know the value of k, we can substitute it back into the original equation to find the first three terms in the expansion in ascending powers of x.

Using n = 3 and k = 1, the binomial expansion becomes:

(1 + x/1)^3 = 1 + 3(x/1) + 3(x/1)^2 + (x/1)^3

Simplifying, we have:

(1 + x)^3 = 1 + 3x + 3x^2 + x^3

Now we can see that the first three terms in the expansion in ascending powers of x are:

1, 3x, 3x^2

To find the coefficients of x and x^2 in the binomial expansion (1+x/k)^n, we can use the formula for the general term in the binomial expansion.

The general term in the binomial expansion is given by:

C(n, r) * (a^(n-r)) * (b^r)

Where:
- C(n, r) represents the binomial coefficient, which is the number of ways to choose r items from a set of n items.
- a and b represent the base terms in the binomial expression.
- n represents the power to which the binomial is raised.
- r represents the power of the first term in the binomial expression.

In this case, a = 1, b = x/k, and the powers of a and b are (n-r) and r, respectively.

(a) To show that the coefficients of x and x^2 are equal, we can equate the general terms for the two powers:

C(n, 1) * (1^(n-1)) * ((x/k)^1) = C(n, 2) * (1^(n-2)) * ((x/k)^2)

Simplifying this equation, we get:

(n * (n-1)!)/((1! * (n-1)!)) * (1^1) * (x^1/k^1) = (n * (n-1) * (n-2)!)/((2! * (n-2)!)) * (1^2) * (x^2/k^2)

Cancelling out the factorials and simplifying further, we have:

n * x/k = (n * (n-1))/2 * (x^2/k^2)

To compare the coefficients of x and x^2, we can ignore the factor of x/k, which leaves us with:

n = (n * (n-1))/2

Multiplying both sides by 2, we get:

2n = n * (n-1)

Expanding the right side of the equation, we have:

2n = n^2 - n

Rearranging the equation, we get:

n^2 - 3n = 0

Factoring out n, we have:

n(n - 3) = 0

Therefore, either n = 0 or n - 3 = 0. Since n is a positive integer, n cannot be 0. Thus, we have:

n - 3 = 0
n = 3

Substituting n = 3 into the equation 2k = n - 1, we can find the value of k:

2k = 3 - 1
2k = 2
k = 1

(b) The value of k is 1.

To find the first three terms in the expansion in ascending powers of x, we can substitute the values of n and k into the binomial expansion formula:

(1+x/k)^n = (1+x/1)^3

Expanding this using the binomial expansion formula:

(1+x)^3 = C(3, 0) * (1^3) * (x^0) + C(3, 1) * (1^2) * (x^1) + C(3, 2) * (1^1) * (x^2) + C(3, 3) * (1^0) * (x^3)

Simplifying this expression, we have:

1 + 3x + 3x^2 + x^3

Therefore, the first three terms in the expansion in ascending powers of x are:

1, 3x, 3x^2