Question

Can someone please help me?

Answer 1

First show it's true for $n=1$. On the left,

$\dfrac1{1\cdot2}=\dfrac12$

On the right,

$\dfrac1{1+1}=\dfrac12$

so the base case $n=1$ is true.

Now assume equality holds for $n=k$, that

$\dfrac1{1\cdot2}+\dfrac1{2\cdot3}+\cdots\dfrac1{k(k+1)}=\dfrac k{k+1}$

We use this assumption to show it also holds for $n=k+1$. By hypothesis,

$\dfrac1{1\cdot2}+\dfrac1{2\cdot3}+\cdots+\dfrac1{k(k+1)}+\dfrac1{(k+1)(k+2)}=\dfrac k{k+1}+\dfrac1{(k+1)(k+2)}$

(the first $k$ terms condense to $\dfrac1{k(k+1)}$)

Combining the fractions gives

$\dfrac{k(k+2)}{(k+1)(k+2)}+\dfrac1{(k+1)(k+2)}=\dfrac{k^2+2k+1}{(k+1)(k+2)}=\dfrac{(k+1)^2}{(k+1)(k+2)}=\dfrac{k+1}{k+2}$

which is what we had to establish, thus proving (by induction) equality for all $n$.

Answer 2

Answer:

im doing the same work but i am working on it ill help you afterwards when i am done .

Step-by-step explanation: