I'm studying in preparation for a Mathematical Analysis I examination and I'm solving past exam exercises. If it's any indicator of difficulty, the exercise is Exercise 1 of 4, part $a$ and graded for 8%, so it's supposed to be pretty easy.
Use induction to prove that $(a+b)^n=a^n+na^{n-1}b+\dfrac{n(n-1)}{2!}a^{n-2}b^2+\dfrac{n(n-1)(n-2)}{3!}a^{n-3}b^3+...+nab^{n-1}+b^n$ for any $a,b\in\mathbb{R}$ and $n\in\mathbb{N}$
I've got the three steps in mind, but I'm failing at the third.
It's obviously true for n=1: $(a+b)^1=a^1+1a^0b$
Let it be true for n=k: $(a+b)^k=a^k+ka^{n k-1}b+\dfrac{k(k-1)}{2!}a^{k-2}b^2+\dfrac{k(k-1)(k-2)}{3!}a^{k-3}b^3+...+kab^{k-1}+b^k$ The third step is obviously the problematic one for me. By expanding the formula for n=k+1, I fail to produce something recognisable that will lead me to something valid like $(a+b)(a+b)^k$
Is there something I'm missing?
Hint: In the third step, try multiplying the RHS of what you got in the second step with $(a+b)$.
That will create two distinct series of sums, say $S_1$ and $S_2$, where $$S_1=a(a^k+ka^{k-1}b+\cdots+kab^{k-1}+b^k)=a^{k+1}+ka^kb+\cdots+ka^2b^{k-1}+ab^k$$ $$S_2=b(a^k+ka^{k-1}b+\cdots+kab^{k-1}+b^k)=a^kb+ka^{k-1}b^2+\cdots+kab^k+b^{k+1}$$ If you leave out the first term in $S_1$ and last term in $S_2$, then its just a matter of using $${x \choose y}+{x \choose y-1}={x+1 \choose y}$$ which can be easily proved.