$$\text { Prove that } \sum _ { j = n } ^ { 2 n - 1 } ( 2 j + 1 ) = 3 n ^ { 2 } \text { whenever } n \text { is a positive integer. }$$

Question

Examlex · Accepted Answer

The Answer is The basis step holds since

\sum _ { j = 1 } ^ { 1 } ( 2 j + 1 ) = 3 = 3 \cdot 1 ^ { 2 }

. For the inductive step assume that

\sum _ { j = k } ^ { 2 k - 1 } ( 2 j + 1 ) = 3 k ^ { 2 }

. It follows that

\sum _ { j = k + 1 } ^ { 2 ( k + 1 ) - 1 } ( 2 j + 1 ) = \sum _ { j = k } ^ { 2 k - 1 } ( 2 j + 1 ) - ( 2 k + 1 ) + ( 4 k + 1 ) + ( 4 k + 3 ) = 3 k ^ { 2 } + 6 k + 3 = 3 ( k + 1 ) ^ { 2 }

. This completes the proof.

Prove that ∑j=n2n−1(2j+1)=3n2 whenever n is a positive integer. \text { Prove that } \sum _ { j = n } ^ { 2 n - 1 } ( 2 j + 1 ) = 3 n ^ { 2 } \text { whenever } n \text { is a positive integer. } Prove that j=n∑2n−1​(2j+1)=3n2 whenever n is a positive integer.