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