use series (in calculus) to prove that 34²=1156 334²=111556 3334²=11115556 33334²=1111155556 3334²=11111155555?
use series (in calculus) to prove that 34²=1156 334²=111556 3334²=11115556 33334²=1111155556 3334²=11111155555
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There is most likely an easier way of doing this problem, and unfortunately I cannot see it, but here is a solution regardless. The numbers on the left hand sides can be written in the form (where n is the number of 3's)
(4 + 30Σ (k = 0 to n - 1) 10^k)² =
(4 + 30[(10^n - 1)/9])² =
(4 + (10/3)(10^n - 1))² =
16 + (80/3)(10^n - 1) + (100/9)(10^(2n) - 2*10^n + 1) =
16 + (240/9)(10^n - 1) + (100/9)(100^n - 1 - 2*10^n + 2) =
16 + (240/9)(10^n - 1) + (1100/99)(100^n - 1) - (200/9)(10^n - 1) =
16 + (40/9)(10^n - 1) + 1100[(100^n - 1)/(100 - 1)] =
16 + 40[(10^n - 1)/(10 - 1)] + 1100Σ (k = 0 to n - 1) 100^k =
16 + 40Σ (k = 0 to n - 1) 10^k + 10*100Σ (k = 0 to n - 1) 100^k + 100Σ (k = 0 to n - 1) 100^k =
16 + 4Σ (k = 1 to n) 10^k + 10Σ (k = 1 to n) 100^k + Σ (k = 1 to n) 100^k =
16 + 4Σ (k = 1 to n) 10^k + Σ (k = 1 to n) 10^(2k + 1) + Σ (k = 1 to n) 10^(2k) =
6 + 10 + 4Σ (k = 1 to n) 10^k + Σ (k = 3, 5, ..., 2n + 1) 10^k + Σ (k = 2, 4, ..., 2n) 10^k =
6 + (4 + 1)Σ (k = 1 to n) 10^k + Σ (k = n + 1 to 2n + 1) 10^k =
6 + 5Σ (k = 1 to n) 10^k + Σ (k = n + 1 to 2n + 1) 10^k
which happens to be the numbers on the right hand sides.