The sum $$\frac{3}{1^2} + \frac{5}{1^2+2^2} + \frac{7}{1^2+2^2+3^2} + \ldots$$ upto 11-terms is:

We consider the general pattern of the terms first.

The first term is $$\dfrac{3}{1^2}$$, the second is $$\dfrac{5}{1^2+2^2}$$, the third is $$\dfrac{7}{1^2+2^2+3^2}$$ and so on.

Observe that for the $$n^{\text{th}}$$ term:

• The numerator is $$3,5,7,\ldots$$ which we can write as $$2n+1$$.

• The denominator is the sum of squares $$1^2+2^2+\ldots+n^2$$.

Hence the $$n^{\text{th}}$$ term can be written as

$$T_n=\dfrac{2n+1}{1^2+2^2+\ldots+n^2}.$$

Now we recall the standard formula for the sum of squares of the first $$n$$ natural numbers:

$$1^2+2^2+\ldots+n^2=\dfrac{n(n+1)(2n+1)}{6}.$$

Substituting this result in the denominator of $$T_n$$, we obtain

$$T_n=\dfrac{2n+1}{\dfrac{n(n+1)(2n+1)}{6}}.$$

Because the factor $$2n+1$$ appears in both numerator and denominator, it cancels, leaving

$$T_n=\dfrac{6}{n(n+1)}.$$

So the required series up to 11 terms is

$$\sum_{n=1}^{11}T_n=\sum_{n=1}^{11}\dfrac{6}{n(n+1)}.$$

Next we simplify the individual fraction $$\dfrac{1}{n(n+1)}$$ using the method of partial fractions.

We write

$$\dfrac{1}{n(n+1)}=\dfrac{A}{n}+\dfrac{B}{n+1}.$$

Multiplying both sides by $$n(n+1)$$, we get

$$1=A(n+1)+Bn.$$

Putting $$n=0$$ gives $$A=1$$, and putting $$n=-1$$ gives $$B=-1$$.

Hence

$$\dfrac{1}{n(n+1)}=\dfrac{1}{n}-\dfrac{1}{n+1}.$$

Using this, we rewrite each term of our series:

$$\dfrac{6}{n(n+1)}=6\left(\dfrac{1}{n}-\dfrac{1}{n+1}\right).$$

Therefore the entire sum becomes

$$\sum_{n=1}^{11}6\left(\dfrac{1}{n}-\dfrac{1}{n+1}\right) =6\sum_{n=1}^{11}\left(\dfrac{1}{n}-\dfrac{1}{n+1}\right).$$

Now we notice that the series inside is telescoping: consecutive negative terms cancel with the next positive terms.

Writing the terms out explicitly, we have

$$\left(\dfrac{1}{1}-\dfrac{1}{2}\right) +\left(\dfrac{1}{2}-\dfrac{1}{3}\right) +\left(\dfrac{1}{3}-\dfrac{1}{4}\right) +\ldots +\left(\dfrac{1}{11}-\dfrac{1}{12}\right).$$

Everything cancels except the first positive fraction $$\dfrac{1}{1}$$ and the very last negative fraction $$-\dfrac{1}{12}$$, so we obtain

$$\sum_{n=1}^{11}\left(\dfrac{1}{n}-\dfrac{1}{n+1}\right)=1-\dfrac{1}{12}=\dfrac{11}{12}.$$

Multiplying by the factor 6 that we extracted earlier gives

$$6\times\dfrac{11}{12}=\dfrac{66}{12}=\dfrac{11}{2}.$$

Hence, the required sum of the first 11 terms of the given series is

$$\dfrac{11}{2}.$$

Hence, the correct answer is Option C.