fix: correct formula for b_N in putnam_like Set_3/B4 grading_scheme.md

[Ashutosh0x]

Fixes #3

Problem

The grading scheme for putnam_like_set3_b4 has an incorrect expression in the last step (the squeeze theorem argument). The formula for $b_N$ used an explicit sum $\sum_{k=1}^{N_n} \frac{k(k+1)(2k+1)}{6}$ which does not correctly represent $\sum_{k=1}^{N_n} a_k$ for this sequence.

Since $b_{N_n} = \frac{\sum_{k=1}^{N_n} a_k}{N_n^\alpha}$, we have $\sum_{k=1}^{N_n} a_k = b_{N_n} \cdot N_n^\alpha$.

Changes

Line 52 — Replaced the incorrect sum expression:

-b_N=\frac{\sum_{k=1}^{N_n} \frac{k(k+1)(2k+1)}6+1+2+\ldots+m}{(N_n+m)^{\alpha}}
+b_N=\frac{b_{N_n}\cdot N_n^\alpha +1+2+\ldots+m}{(N_n+m)^{\alpha}}

Line 56 — Simplified the squeeze theorem bounds by removing incorrect intermediate equalities:

-b_{N_n}\cdot\frac{N_n^{\alpha}}{N_{n+1}^{\alpha}}=\frac{\sum_{k=1}^{N_n} \frac{k(k+1)(2k+1)}6}{N_{n+1}^{\alpha}}<b_N<\frac{\sum_{k=1}^{N_{n+1}} \frac{k(k+1)(2k+1)}6}{N_n^{\alpha}}=b_{N_{n+1}}\cdot\frac{N_{n+1}^{\alpha}}{N_{n}^{\alpha}}.
+b_{N_n}\cdot\frac{N_n^{\alpha}}{N_{n+1}^{\alpha}}<b_N<b_{N_{n+1}}\cdot\frac{N_{n+1}^{\alpha}}{N_{n}^{\alpha}}.

As noted in the issue, these corrections do not impact the final answer or the overall idea of the solution.