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Question Number 121655 by TITA last updated on 10/Nov/20

Commented by Dwaipayan Shikari last updated on 10/Nov/20

$$1=1^2$$$$1+3=2^2$$$$1+3+5=3^2$$$$1+3+5+7=4^2$$$$1+3+5+7+9=5^2..$$$$\underset{n=1}{\sum ^n}(2n-1)=n^2$$$$n^2⩾4000n⩾63.2n_{min}=64$$

Commented by TITA last updated on 10/Nov/20

$$thankssir$$

Answered by 675480065 last updated on 10/Nov/20

$$\sum _{\mathrm{k}=1}^{\mathrm{n}}(2\mathrm{k}-1)={\mathrm{n}}^2$$$$\mathrm{Use}\mathrm{mathematical}\mathrm{induction}$$

Answered by TANMAY PANACEA last updated on 11/Nov/20

$$Directmethod$$$$S=1+3+5+7..+(2n-1)$$$$=\frac{n}{2}[2\times 1+(n-1)\times 2]\to S=\frac{n}{2}(2a+(n-1)d]$$$$=n^2$$

Commented by bemath last updated on 11/Nov/20

$$itshouldbe1+3+5+7+...+(2n-1)$$$$sir$$

Commented by TANMAY PANACEA last updated on 11/Nov/20

$$yesyesthankyou$$

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