Prove that $$\sum _ { j = n } ^ { 2 n - 1 } ( 2 j + 1 ) = 3 n ^ { 2 } \text { for all positive integers } n$$

Name: Prove that \[\sum _ { j = n } ^ { 2 n - 1 } ( 2 j + 1 ) = 3 n ^ { 2 } \text { for all positive integers } n\]
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Question

Quizplus · Accepted Answer

The Answer is The basis case holds since

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

. Now assume that

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

for some

k

. 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 }

.

Prove That ∑j=n2n−1(2j+1)=3n2 for all positive integers n\sum _ { j = n } ^ { 2 n - 1 } ( 2 j + 1 ) = 3 n ^ { 2 } \text { for all positive integers } nj=n∑2n−1​(2j+1)=3n2 for all positive integers n

Prove That $\sum _ { j = n } ^ { 2 n - 1 } ( 2 j + 1 ) = 3 n ^ { 2 } \text { for all positive integers } n$