Question

The number of terms in the expansion of (1+ x3 + x4)8 is​

Answer 1

(x+a)47−(x−a)47

When we expand the above equation using binomial expansion

(x+y)n=(k=0∑nnCkxkyn−k)

So the above equation becomes

(x+a)47=(k=0∑4747Ckxka47−k)

(x−a)47=(k=0∑4747Ckxk(−a)47−k)

(x+a)47⇒There are 48 terms in the expansion and all are positive

(x−a)47⇒There are 48 terms in the expansion

The terms with odd powers of a will be cancelled and those with even powers of a will add up.

24 terms will be positive and 24 negative in the expansion of (x−a)47

48 terms positive-[24 terms negative and 24 terms positive]

=48termspositive+24termsnegative+24termspositive

=24 terms

Answer 2

Step-by-step explanation:

Since all coefficients are positive, there is no concern about terms possibly canceling out.

It is easy to show that (1+x3+x5)3 has 10 terms. You can either list them out, or use (52)=10.

Proceed by induction, to show that if n≥3, then fn(x)=(1+x3+x5)n has 5n−5 terms.

Hint: Going from k to k+1, all coefficients which existed in fk will also exist in fk+1.

Hint: Show that fk+1 has 5 additional terms. Which 5 are these?

Hint: 5(k+1) is one of the additional terms. Do you see why? What other additional terms are there?