Exercise 7, § 6 Algebra, IV, p. 99 With the notation of Exercise 6 show that ∑aλ1λ2…λn=−1+k∑j=0k/(n+1)(−1)jk−jn(k−jnj)2k−j(n+1) \sum a_{\lambda_1 \lambda_2 \ldots \lambda_n} = -1 + k \sum_{j=0}^{k/(n+1)} \frac{(-1)^j}{k - jn} \binom{k - jn}{j} 2^{k - j(n+1)} ∑aλ1λ2…λn=−1+kj=0∑k/(n+1)k−jn(−1)j(jk−jn)2k−j(n+1) Solution No solution has been written yet.