∑i=1n2i+i=3n\sum_{i=1}^{n}{2^i+i=3n}i=1∑n2i+i=3n
asked by guest on Feb 02, 2025 at 10:22 am
I wasn't able to parse your question, but the HE.NET team is hard at work making me smarter.
Below are 10 similar questions that I was able to parse.
∑i=1n2i∗n\sum_{i=1}^{n}{2^i*n}i=1∑n2i∗n
∑i=1n2i+1\sum_{i=1}^{n}{2^i+1}i=1∑n2i+1
∑i=0ne−2i/n\sum_{i=0}^{n}{e^-2i/n}i=0∑ne−2i/n
∑i=1nai\sum_{i=1}^{n}{a^i}i=1∑nai
∑i=0n2i\sum_{i=0}^{n}{2^i}i=0∑n2i
∑i=0nni\sum_{i=0}^{n}{n^i}i=0∑nni
∑i=0nri\sum_{i=0}^{n}{r^i}i=0∑nri
∑i=13δii\sum_{i=1}^{3}{δii}i=1∑3δii
∑i=0n9i\sum_{i=0}^{n}{9^i}i=0∑n9i
