Forgot password? New user? Sign up
Existing user? Log in
12+22+32+⋯+k2=(N−1)N(N+1)24,1^2+2^2+3^2+\cdots+k^2 = \frac{(N-1)N(N+1)}{24},12+22+32+⋯+k2=24(N−1)N(N+1), where N=2k+1N = 2k+1N=2k+1.
Now consider the sum (12)2+(112)2+(212)2+⋯+(k−12)2.(\tfrac12)^2+(1\tfrac12)^2+(2\tfrac12)^2+\cdots+(k-\tfrac12)^2.(21)2+(121)2+(221)2+⋯+(k−21)2. Can this sum also be written in the form (N−1)N(N+1)24 ?\frac{(N-1)N(N+1)}{24}\ ?24(N−1)N(N+1) ? If so, what is the value of NNN?
Problem Loading...
Note Loading...
Set Loading...