Mathematics, 11.03.2021 04:00 ponyolovezsosuk1074
Use mathematical induction to prove the statement is true for all positive integers n. 1^2 + 3^2 + 5^2 + ... + (2n-1)^2 = (n(2n-1)(2n+1))/3)
Answers: 1
.
:
, under the assumption that 
is true:![(1^{2}+2^{2}+3^{2}+...+k^{2}) + [2\cdot (k+1)-1]^{2} = \frac{(k+1)\cdot [2\cdot (k+1)-1]\cdot [2\cdot (k+1)+1]}{3}](/tpl/images/1186/3827/e14f4.png)
![\frac{k\cdot (2\cdot k -1)\cdot (2\cdot k +1)}{3} +[2\cdot (k+1)-1]^{2} = \frac{(k+1)\cdot [2\cdot (k+1)-1]\cdot [2\cdot (k+1)+1]}{3}](/tpl/images/1186/3827/41fa4.png)
![\frac{k\cdot (2\cdot k -1)\cdot (2\cdot k +1)+3\cdot [2\cdot (k+1)-1]^{2}}{3} = \frac{(k+1)\cdot [2\cdot (k+1)-1]\cdot [2\cdot (k+1)+1]}{3}](/tpl/images/1186/3827/88cc3.png)
![(2\cdot k +1)\cdot [k\cdot (2\cdot k -1)+3\cdot (2\cdot k +1)] = (k+1) \cdot (2\cdot k +1)\cdot (2\cdot k +3)](/tpl/images/1186/3827/9bdd1.png)
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Use mathematical induction to prove the statement is true for all positive integers n. 1^2 + 3^2 + 5...
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