1. Interpret the limit carefully

We need to evaluate

$$\lim_{n\to\infty} \frac{1^a+2^a+\cdots+n^a}{(n+1)^{a-1}\,[(na+1)+(na+2)+\cdots+(na+n)]}=\frac1{60}$$

with $a\in\mathbb R$ and $a\neq -1$.

We must determine which options satisfy this.

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2. Simplify the sum in the denominator

Consider

$$(na+1)+(na+2)+\cdots+(na+n).$$

This is a sum of $n$ terms:

$$= n(na) + (1+2+\cdots+n)$$ $$= n^2a + \frac{n(n+1)}{2}.$$

So the denominator becomes

$$(n+1)^{a-1}\left(n^2a+\frac{n(n+1)}2\right).$$

For large $n$,

$$(n+1)^{a-1}\sim n^{a-1},$$

and

$$n^2a+\frac{n(n+1)}2 \sim n^2\left(a+\frac12\right).$$

Hence the whole denominator behaves like

$$n^{a-1}\cdot n^2\left(a+\frac12\right)=n^{a+1}\left(a+\frac12\right).$$

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3. Asymptotic form of the numerator

For $a>-1$,

$$1^a+2^a+\cdots+n^a \sim \frac{n^{a+1}}{a+1}.$$

Therefore the limit becomes

$$\lim_{n\to\infty} \frac{\frac{n^{a+1}}{a+1}}{n^{a+1}\left(a+\frac12\right)} = \frac{1}{(a+1)\left(a+\frac12\right)}.$$

So we must have

$$\frac{1}{(a+1)\left(a+\frac12\right)}=\frac1{60}.$$

Thus

$$(a+1)\left(a+\frac12\right)=60.$$

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4. Solve for $a$

Expand:

$$(a+1)\left(a+\frac12\right)=a^2+\frac32 a+\frac12.$$

So

$$a^2+\frac32 a+\frac12=60$$ $$a^2+\frac32 a-\frac{119}{2}=0.$$

Multiply by $2$:

$$2a^2+3a-119=0.$$

Now factor:

$$2a^2+17a-14a-119=0$$ $$( a-7)(2a+17)=0.$$

Hence

$$a=7 \quad \text{or} \quad a=-\frac{17}{2}.$$

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5. Check validity of both roots

Our asymptotic formula

$$\sum_{k=1}^n k^a \sim \frac{n^{a+1}}{a+1}$$

was used for $a>-1$.

So:

• $a=7$ is valid.
• $a=-\frac{17}{2}<-1$ is not valid for that asymptotic formula.

Now check $a=-\frac{17}{2}$ directly.

If $a<-1$, then

$$\sum_{k=1}^n k^a$$

converges to a finite positive constant as $n\to\infty$.

But the denominator behaves as

$$(n+1)^{a-1}\left(n^2a+\frac{n(n+1)}2\right) \sim n^{a+1}\left(a+\frac12\right).$$

For $a=-\frac{17}{2}$,

$$a+1=-\frac{15}{2},$$

so

$$n^{a+1}=n^{-15/2}\to 0.$$

Hence the denominator tends to $0$ (not a finite nonzero number), while the numerator tends to a positive finite constant. Therefore the quotient does not approach $\frac1{60}$.

So $a=-\frac{17}{2}$ is rejected.

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6. Evaluate options

• A: $5$

  $$\frac{1}{(5+1)(5+1/2)}=\frac{1}{6\cdot 11/2}=\frac1{33}\neq \frac1{60}$$

  Not correct.

• B: $7$

  $$\frac{1}{(7+1)(7+1/2)}=\frac{1}{8\cdot 15/2}=\frac1{60}$$

  Correct.

• C: $-\frac{15}{2}$ Here $a<-1$, so numerator converges, denominator $\to 0$; limit is not $\frac1{60}$. Not correct.

• D: $-\frac{17}{2}$ Same issue; not correct.

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7. Final answer

The only correct option is

$$\boxed{7}.$$