Finding Binomial Terms | CAIE 9709 Pure Mathematics 1 Notes

Pure Mathematics 1

Mathematics9709

Find the required power before calculating

If only one coefficient is requested, you do not need a full expansion. The term made by selecting the second part $k$ times is

T_{k+1}=\binom nk a^{n-k}b^k

$T$ labels a term in this expansion. The index is $k+1$ because the first term uses $k=0$ . The fourth term uses $k=3$ , not 4.

Worked example

Find the coefficient of $x^4$ in $(3+2x)^6$ .

The second part contributes $x^k$ , so use $k=4$ . Calculate the count with nCr, then the rest of the term:

\binom64 3^2(2x)^4=2160x^4

The requested coefficient is 2160, not $2160x^4$ . If asked for the term, include the power.

With $(2-x^2)^7$ , the second part contributes $x^{2k}$ , so the required power is not automatically $k$ .

When both parts contain x

Use $x^p x^q=x^{p+q}$ and $1/x^q=x^{-q}$ . First simplify the total power of $x$ , then solve for $k$ .

Worked example

Find the term independent of $x$ in $(x^2+2/x)^6$ , where $x\ne0$ .

“Independent of $x$ ” means the constant term: its power is zero.

T_{k+1}=\binom6k(x^2)^{6-k}(2/x)^k

Multiply the powers in $(x^2)^{6-k}$ , then add the negative power from $1/x^k$ :

x^{2(6-k)}x^{-k}=x^{12-3k}

T_{k+1}=\binom6k2^k x^{12-3k}

Set $12-3k=0$ , so $k=4$ . The term is $\binom64 2^4=240$ .

A valid choice count must be an integer from 0 to $n$ . Here a requested $x^2$ term would need $k=10/3$ , so that term is absent and its coefficient is zero.

Now you try

Your turnIndependent practice

(a) Find the coefficient of $x^6$ in $(2-x^2)^7$ .

(b) Find the term independent of $x$ in $(2x-1/x)^4$ , for $x\ne0$ .

Show worked answer

(a) $2k=6$ , so $k=3$ . The term is $\binom73 2^4(-x^2)^3=-560x^6$ ; the coefficient is −560.

(b) The general term is $\binom4k2^{4-k}(-1)^k x^{4-2k}$ . Set $4-2k=0$ to get $k=2$ . The constant is $\binom42 2^2=24$ .

Find the required power before calculating

If only one coefficient is requested, you do not need a full expansion. The term made by selecting the second part $k$ times is

T_{k+1}=\binom nk a^{n-k}b^k

$T$ labels a term in this expansion. The index is $k+1$ because the first term uses $k=0$ . The fourth term uses $k=3$ , not 4.

Worked example

Find the coefficient of $x^4$ in $(3+2x)^6$ .

The second part contributes $x^k$ , so use $k=4$ . Calculate the count with nCr, then the rest of the term:

\binom64 3^2(2x)^4=2160x^4

The requested coefficient is 2160, not $2160x^4$ . If asked for the term, include the power.

With $(2-x^2)^7$ , the second part contributes $x^{2k}$ , so the required power is not automatically $k$ .

When both parts contain x

Use $x^p x^q=x^{p+q}$ and $1/x^q=x^{-q}$ . First simplify the total power of $x$ , then solve for $k$ .

Worked example

Find the term independent of $x$ in $(x^2+2/x)^6$ , where $x\ne0$ .

“Independent of $x$ ” means the constant term: its power is zero.

T_{k+1}=\binom6k(x^2)^{6-k}(2/x)^k

Multiply the powers in $(x^2)^{6-k}$ , then add the negative power from $1/x^k$ :

x^{2(6-k)}x^{-k}=x^{12-3k}

T_{k+1}=\binom6k2^k x^{12-3k}

Set $12-3k=0$ , so $k=4$ . The term is $\binom64 2^4=240$ .

A valid choice count must be an integer from 0 to $n$ . Here a requested $x^2$ term would need $k=10/3$ , so that term is absent and its coefficient is zero.

Now you try

Your turnIndependent practice

(a) Find the coefficient of $x^6$ in $(2-x^2)^7$ .

(b) Find the term independent of $x$ in $(2x-1/x)^4$ , for $x\ne0$ .

Show worked answer

(a) $2k=6$ , so $k=3$ . The term is $\binom73 2^4(-x^2)^3=-560x^6$ ; the coefficient is −560.

(b) The general term is $\binom4k2^{4-k}(-1)^k x^{4-2k}$ . Set $4-2k=0$ to get $k=2$ . The constant is $\binom42 2^2=24$ .