Math, asked by pournimakharat017, 3 months ago

find the sum of first n natural number​


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Answers

Answered by NikhilAhire
2

Answer:

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Step-by-step explanation:

S=

2

n(n+1)

⇒231=

2

n(n+1)

⇒n

2

+n−462=0

⇒n

2

+22n−21n−462=0

⇒n(n+22)−21(n+22)=0

⇒(n+22)(n−21)=0

⇒n=−22 or n=21

∴n=21 as n=−22 is not a natural number.


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Answered by anshikaroy3456
0

We prove the formula 1+ 2+ ... + n = n(n+1) / 2, for n a natural number.

There is a simple applet showing the essence of the inductive proof of this result. To run this applet, you first enter the number n you wish to have illustrated; space limitations require 0<n<11. Then push the [Next] button to step through the stages of the proof.

The base case shown by the applet is n=1, although on the proof pages the base case is n=0; this is just because there is nothing to show when n=0.

Assuming the result for n means we know how to sum half of an nx(n+1) rectangle having rows with 1, 2, ..., n red dots, respectively. Now we add a new row with all black dots, and then one more red dot to each row. The result is another figure of the same form, but with the parameter n+1 instead of n


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