sum_(k=1)^(2 n) (-1)^k 2^(k - 1) binomial(2 n, k) = 0

asked by guest
on Jan 15, 2025 at 12:57 pm



You asked:

Investigate the equation: ∑k=12n(−1)k⋅2k−1⋅C⁡(2n,k)=0\sum_{k = 1}^{2 n} {{\left( -1 \right)}^{k} \cdot {2}^{k - 1} \cdot \operatorname{C}(2 n, k)} = 0.

MathBot Answer:

No solution.

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