What is the sum of squares of the first 2006 numbers in the sequence of 1212212212222?

What is the sum of squares of the first 2006 numbers in the sequence of 1212212212222?

If there are n ones, then n + 1 + 2 + 3 + +n ≥2006
N (n + 3) ≥ 4012
Because 61 * 64 = 3904, 62 * 65 = 4030, the last term of n = 62 has 62 - (4030 -- 4012) = 44 2, so the final result is 62 * 1 ^ 2 + (1 + 2 + 3 +...) +61+44）*2^2=7802

If a sequence diverges and a sequence converges, does the sum of their squares converge or diverge

It may converge or diverge
For example, if an = (- 1) ^ n and BN = 1, the sequence {an} diverges and the sequence {BN} converges, while the sequence {an ^ 2 + BN ^ 2} = the sequence {2} converges
Another example is an = n, BN = 1, sequence {an} diverges, sequence {BN} converges, and sequence {an ^ 2 + BN ^ 2} = sequence {1 + n ^ 2} diverges

If the common ratio of an infinite equal ratio sequence satisfies the absolute value Q < 1, the sum of its terms is equal to 6, the sum of its squares is equal to 18, what is q?

His sum of terms = A1 / (1-Q) = 6
square
a²/(1-q)²=36
Square, then the first term is A1 & sup2; and the common ratio is Q & sup2;
So sum = A1 & sup2; / (1-Q & sup2;) = 18
be divided by
(1-q²)/(1-q)²=36/18
(1+q)(1-q)/(1-q)²=2
(1+q)/(1-q)=2
1+q=2-2q
q=1/3