Prove using mathematical induction that for all n - 11-4-7-3n-2-n3n-1-2

For any integer n ≥ 1,P_n be the statement that 1+4+7+..+(3n 2)=n(3n 1)/2 Base case: The statement P_1 says that 1=1(3 1)/2. Which is true. Inductive step. ...

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Standard XI

Mathematics

Proof by mathematical induction

Prove using m...

Question

Prove using mathematical induction that for all $n \geq 1$

1+4+7+..+(3n-2)= $\frac{n (3 n - 1)}{2}$

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For any integer $n \geq 1$ , $P_{n}$ be the statement that

1+4+7+..+(3n-2)= $\frac{n (3 n - 1)}{2}$

$B a s e c a s e - --------- -$ : The statement $P_{1}$ says that

$1 = \frac{1 (3 - 1)}{2}$ .

Which is true.

Inductive step. Fix $k \geq 1$ , and supoose that $P_{k}$ holds, that is,

$1 + 4 + 7 + . . . + (3 k - 2) = \frac{k (3 k - 1)}{2}$ .

It remains to show that $p_{k + 1}$ holds, that is,

$1 + 4 + 7 + . . . + (3 (k + 1) - 2) = \frac{(k + 1) (3 (k + 1) - 1)}{2}$ .

$1 + 4 + 7 + . . . + (3 (k + 1) - 2) = 1 + 4 + 7 + . . . + (3 (k + 1) - 2)$

= $1 + 4 + 7 + . . . + (3 k + 1)$

= $1 + 4 + 7 + . . . + (3 k - 2) + (3 k + 1)$

= $\frac{k (3 k - 1)}{2} + (3 k + 1)$

= $\frac{k (3 k - 1) + 2 (3 k + 1)}{2}$

= $\frac{3 k^{2} - k + 6 k + 2}{2}$

= $\frac{3 k^{2} + 5 k + 2}{2}$

= $\frac{(k + 1) (3 k + 2)}{2}$

= $\frac{(k + 1) (3 (k + 1) - 1)}{2}$

$∴ P_{k + 1} h o l d s$ .

thus, by the principle of mathematical induction for all $n \geq 1$ , $P_{n}$ holds.

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