Question

Let
\(𝑆_𝑛 =∑_{k=1}^n\frac{1+k2^k}{4^{k-1}},n=1,2,...\)
Then, lim𝑛_→∞ 𝑆_𝑛 equals (round off to two decimal places)

Answer 1

Correct Answer: 9.32

Problem (as written) 

Let $$S_n=\sum_{k=1}^{n}\; \text{(term unclear from the typing)}\qquad n=1,2,\dots$$ You gave the numeric value (limit) as 9.32.

Note about the statement

The typed summand appears garbled (e.g. the line shows fragments like 1 + k 2 k 4 k − 1). There are several natural ways to read that expression; below I evaluate the limits for the most plausible interpretations so you can see which (if any) equals 9.32.

Working — three plausible interpretations

Interpretation A: $$S_n=\sum_{k=1}^{n}\frac{1+k^2}{4^{\,k-1}}$$ This is a simple power-series type sum. Using standard summation (or direct evaluation), $$S_{\infty}=\sum_{k=1}^{\infty}\frac{1+k^2}{4^{\,k-1}} \approx 4.2962962963.$$ Rounded to two decimals: $4.30$.

---

Interpretation B: $$S_n=\sum_{k=1}^{n}\frac{(1+k^2)\,k}{4^{\,k-1}}$$ (i.e. an extra factor $k$ in the numerator). Direct summation gives $$S_{\infty}=\sum_{k=1}^{\infty}\frac{(1+k^2)\,k}{4^{\,k-1}} \approx 8.2962962963.$$ Rounded to two decimals: $8.30$.

---

Interpretation C: $$S_n=\sum_{k=1}^{n}\frac{(1+k^2)k^2}{4^{\,k-1}}$$ (i.e. factor $k^2$ in numerator). This sums to $$S_{\infty}\approx 21.7283950617.$$ Rounded to two decimals: $21.73$.

Conclusion

None of the three most natural readings above gives the value 9.32. The values (rounded to two decimals) are:

• Interpretation A: 4.30
• Interpretation B: 8.30
• Interpretation C: 21.73

Of these, Interpretation B (8.30) is the closest to 9.32 but still differs substantially.