5
$\begingroup$

$$ \sum_{k=1}^n \frac {k+2}{k(k+1)2^{k+1}} = \frac{1}{2} \ - \frac1{(n+1)2^{n+1}} $$

Base Case:

I did $n = 1$, so..

LHS-

$$\sum_{k=1}^n \frac {k+2}{k(k+1)2^{k+1}} = \frac3{8}$$

RHS-

$$\frac{1}{2} \ - \frac1{(n+1)2^{n+1}} \ = \frac3{8}$$

so LHS = RHS

Inductive case-

LHS for $n+1$

$$\sum_{k=1}^{n+1} \frac {k+2}{k(k+1)2^{k+1}} +\frac {n+3}{(n+1)(n+2)2^{n+2}}$$

and then I think that you can use inductive hypothesis to change it to the form of $$ \frac{1}{2} \ - \frac1{(n+1)2^{n+1}} +\frac {n+3}{(n+1)(n+2)2^{n+2}} $$

and then I broke up $\frac {n+3}{(n+1)(n+2)2^{n+2}}$ into

$$\frac{2(n+2)-(n+1)}{(n+1)(n+2)2^{n+2}}$$

$$=$$

$$\frac{2}{(n+1)2^{n+2}} - \frac{1}{(n+2)2^{n+2}}$$

$$=$$

$$\frac{1}{(n+1)2^{n+1}} - \frac{1}{(n+2)2^{n+2}}$$

then put it back in with the rest of the equation, bringing me to

$$\frac{1}2 -\frac {1}{(n+1)2^{n+1}} +\frac{1}{(n+1)2^{n+1}} - \frac{1}{(n+2)2^{n+2}}$$

then

$$\frac{1}2 -\frac{2}{(n+1)2^{n+1}} - \frac{1}{(n+2)2^{n+2}}$$

and

$$\frac{1}2 -\frac{1}{(n+1)2^{n}} - \frac{1}{(n+2)2^{n+2}}$$

$$\frac{1}2 -\frac{(n+2)2^{n+2} - (n+1)2^{n}}{(n+1)(n+2)2^{2n+2}} $$

which I think simplifies down to this after factoring out a $2^{n}$ from the numerator?

$$\frac{1}2 -\frac{2^{n}((n+2)2^{2} - (n+1))}{(n+1)(n+2)2^{2n+2}} $$

canceling out $2^{n}$

$$\frac{1}2 -\frac{(3n-7)}{(n+1)(n+2)2^{n+2}} $$

and I'm stuck, please help!

$\endgroup$
4
$\begingroup$

Your error is just after the sixth step from the bottom:

$$\frac{1}2 -\frac {1}{(n+1)2^{n+1}} +\frac{1}{(n+1)2^{n+1}} - \frac{1}{(n+2)2^{n+2}}=\frac{1}2 -\frac{1}{(n+2)2^{n+2}}$$

Then you are done.

You accidentally added the two middle terms instead of subtracting.

$\endgroup$
2
$\begingroup$

Using a telescoping sum, we get $$ \begin{align} \sum_{k=1}^n\frac{k+2}{k(k+1)2^{k+1}} &=\sum_{k=1}^n\left(\frac1{k2^k}-\frac1{(k+1)2^{k+1}}\right)\\ &=\sum_{k=1}^n\frac1{k2^k}-\sum_{k=2}^{n+1}\frac1{k2^k}\\ &=\frac12-\frac1{(n+1)2^{n+1}} \end{align} $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.