If the sum of the first 10 terms of the series (1)/(1+1^4 4)+(2)/(1+2^4 4)+(3)/(1+3^4 4)+(…

Question

If the sum of the first 10 terms of the series $\frac{1}{1+1^4 	imes 4}+\frac{2}{1+2^4 	imes 4}+\frac{3}{1+3^4 	imes 4}+\frac{4}{1+4^4 	imes 4}+\ldots \ldots$. is $\frac{m}{n}, \operatorname{gcd}(m, n)=1$, then $m+n$ is equal to :

Accepted Answer

The given series is

$\frac{1}{1+4\cdot {1}^{4}}+\frac{2}{1+4\cdot {2}^{4}}+\frac{3}{1+4\cdot {3}^{4}}+\cdots$

So the general term is

$\frac{r}{1+4{r}^{4}}$

We need the sum of the first $10$ terms:

${S}_{10}=\sum _{10}^{r=1}\frac{r}{1+4{r}^{4}}$

Now factor the denominator:

$1+4{r}^{4}=(2{r}^{2}-2r+1)(2{r}^{2}+2r+1)$

So,

$\frac{r}{1+4{r}^{4}}=\frac{r}{(2{r}^{2}-2r+1)(2{r}^{2}+2r+1)}$

We now try to split it:

$\frac{r}{(2{r}^{2}-2r+1)(2{r}^{2}+2r+1)}=\frac{1}{4}(\frac{1}{2{r}^{2}-2r+1}-\frac{1}{2{r}^{2}+2r+1})$

Check:

$\frac{1}{4}(\frac{1}{2{r}^{2}-2r+1}-\frac{1}{2{r}^{2}+2r+1})$

Taking LCM,

$=\frac{1}{4}\cdot \frac{(2{r}^{2}+2r+1)-(2{r}^{2}-2r+1)}{(2{r}^{2}-2r+1)(2{r}^{2}+2r+1)}$

$=\frac{1}{4}\cdot \frac{4r}{(2{r}^{2}-2r+1)(2{r}^{2}+2r+1)}$

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

Correct.

Also note that

$2{r}^{2}+2r+1=2(r+1{)}^{2}-2(r+1)+1$

So if we define

${a}_{r}=2{r}^{2}-2r+1$

then

$2{r}^{2}+2r+1={a}_{r+1}$

Hence,

$\frac{r}{1+4{r}^{4}}=\frac{1}{4}(\frac{1}{{a}_{r}}-\frac{1}{{a}_{r+1}})$

Therefore,

${S}_{10}=\frac{1}{4}\sum _{10}^{r=1}(\frac{1}{{a}_{r}}-\frac{1}{{a}_{r+1}})$

This is a telescoping series:

${S}_{10}=\frac{1}{4}(\frac{1}{{a}_{1}}-\frac{1}{{a}_{11}})$

Now compute:

${a}_{1}=2(1{)}^{2}-2(1)+1=1$

and

${a}_{11}=2(11{)}^{2}-2(11)+1=242-22+1=221$

So,

${S}_{10}=\frac{1}{4}(1-\frac{1}{221})$

$=\frac{1}{4}\cdot \frac{220}{221}$

$=\frac{55}{221}$

Thus,

$m=55,\, n=221$

So,

$m+n=55+221=276$

Hence, the correct answer is:

$276$

So, Option C is correct.

Solutions · 1