Using binomial coefficients in a seventh power expansion

Key concept: This is a classic binomial theorem problem where comparing coefficients gives a clean system of equations.

Step by step

Step 1: Match the coefficients

In the expansion of $(x+1)^n$, the coefficients are binomial coefficients:

$\binom{n}{0},\binom{n}{1},\binom{n}{2},\binom{n}{3},\dots$

We are told that

$(x+1)^n=x^7+ax^6+bx^5+35x^4+\dots+1.$

Since the highest power is $x^7$, we must have $n=7$.

Now identify the coefficients:

$a=\binom{7}{1}=7,$ $b=\binom{7}{2}=21,$ $\binom{7}{3}=35,$ which confirms part (a):

$\boxed{b=21}.$

Step 2: Use the condition on the terms

The terms are:

• first term: $x^7$
• second term: $7x^6$
• third term: $21x^5$
• fourth term: $35x^4$

The problem states that the third term is the mean of the second and fourth terms. So

$21x^5=\frac{7x^6+35x^4}{2}.$

Multiply by 2:

$42x^5=7x^6+35x^4.$

Factor out $7x^4$:

$6x= x^2+5.$

So we get the quadratic equation

$x^2-6x+5=0.$

Factor:

$(x-1)(x-5)=0.$

Therefore,

$\boxed{x=1 \text{ or } x=5}.$

Step 3: Check the condition $x\ne 0$

Both values are valid, and neither is zero, so both are acceptable solutions.

Key exam idea

The coefficient pattern in a binomial expansion is controlled by combinations. Once the power is identified as 7, the rest of the coefficients come directly from Pascal's triangle or $\binom{7}{r}$ values. The term relation then turns into an equation in $x$, which is usually the fastest path to the answer.

Pitfall alert

A common mistake is to treat the coefficient $35$ as if it were unrelated to $n$ and try guessing the power from the terms alone. The correct first step is to recognize that the leading term $x^7$ forces $n=7$. Another trap is to forget that the coefficients multiply powers of $x$, so the mean condition must be applied to the full terms, not just to the numerical coefficients. Finally, do not divide by $x^4$ too early without checking that $x\ne 0$; here it is allowed, but that assumption should be stated explicitly.

Try different conditions

If the expansion were $(x+1)^8=x^8+ax^7+bx^6+\cdots$, then the same method would use $a=8$ and $b=28$. If the condition changed to "the third term is twice the fourth term," then you would set $21x^5=2\cdot 35x^4$ and solve directly for $x$. If the base were $(2x+1)^7$, the coefficients would still come from binomial coefficients, but each term would include powers of 2 as well, so the comparison equation would change.

Further reading

binomial coefficients, Pascal's triangle, general term