The Binomial Theorem lets us expand (a+b)n as a sum of terms (rn)an−rbr. For (3x+x1)5, we get six terms. After simplifying powers of x and constants, the final expansion is 243x5+815x3+2710x+9x10+3x35+x51.
The Binomial Theorem is the natural tool here. When you see something like (a+b)n, it’s a compact way of writing a sum where each term picks a a certain number of times and b the rest. The coefficients come from combinations — (rn) — which count how many ways you can choose r copies of b (and n−r copies of a).
In our case, a=3x, b=x1, and n=5. The expansion will have 6 terms (from r=0 to 5). Each term will involve a power of x that changes with r, and a constant coefficient that we simplify.
Let’s work through it.
- Write the general term.
The r-th term (starting with r=0) is
Tr+1=(r5)(3x)5−r(x1)r.
- Simplify the powers of x.
(3x)5−r=35−rx5−r and (x1)r=xr1.
Multiplying:
Tr+1=(r5)⋅35−rx5−r⋅xr1=(r5)⋅35−rx5−2r.
Notice the exponent of x is 5−2r. As r goes from 0 to 5, this exponent decreases by 2 each step: 5,3,1,−1,−3,−5. So the powers of x will be x5,x3,x1,x−1,x−3,x−5 — that is, x5,x3,x,x1,x31,x51.
-
Compute each term one by one.
We’ll need (05)=1, (15)=5, (25)=10, (35)=10, (45)=5, (55)=1.
-
r=0:
T1=1⋅35x5=243x5.
-
r=1:
T2=5⋅34x3=5⋅81x3=815x3.
-
r=2:
T3=10⋅33x1=10⋅27x=2710x.
-
r=3:
T4=10⋅32x−1=10⋅9x1=9x10.
-
r=4: …