FMM11 - Revision history

Admin at 22:36, 16 November 2019

2019-11-16T22:36:27Z

← Older revision		Revision as of 22:36, 16 November 2019
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	<math>		<math>
	1 - 4 + 9 - 16 + ... + (-1)^(n-1) n^2 = (-1)^{n-1} \frac{n(n+1)}{2}		1 - 4 + 9 - 16 + ... + (-1)^{n-1} n^2 = (-1)^{n-1} \frac{n(n+1)}{2}
	</math>		</math>

Admin at 22:36, 16 November 2019

2019-11-16T22:36:04Z

← Older revision		Revision as of 22:36, 16 November 2019
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	The general law is:		The general law is:

	1 - 4 + 9 - 16 + ... + (-1)^(n-1) n^2		<math>
	=		1 - 4 + 9 - 16 + ... + (-1)^(n-1) n^2 = (-1)^{n-1} \frac{n(n+1)}{2}
	(-1)^(n-1~~) (~~n(n+1))/2		</math>

	<math>		<math>
	\sum_{k=1}^n (-1)^{k-1} k^2 = (-1)^{n-1} \frac{(n(n+1))}{2}		\sum_{k=1}^n (-1)^{k-1} k^2 = (-1)^{n-1} \frac{n(n+1)}{2}
	</math>		</math>

Line 29:		Line 29:

	<math>		<math>
	\sum_{k=1}^{n+1} (-1)^{k-1} k^2 = (-1)^{n} \frac{((n+1)(n+2))}{2}		\sum_{k=1}^{n+1} (-1)^{k-1} k^2 = (-1)^{n} \frac{(n+1)(n+2)}{2}
	</math>		</math>

	If you split the left side into the sum from k=1 to n, (which can be reduced to a single term by using the identity above for the case of n, which you assume is true), and the last n+1 term written explicitly, you can do some algebraic manipulation to show it's equivalent to the right side.		If you split the left side into the sum from k=1 to n, (which can be reduced to a single term by using the identity above for the case of n, which you assume is true), and the last n+1 term written explicitly, you can do some algebraic manipulation to show it's equivalent to the right side.
	}}		}}

Admin: Created page with "{{FMM |title=Inferring Rule from Sequence |problem=

              1 = 1          1 - 4 = -(1 + 2)      1 - 4 + 9 = 1 + 2 + 3 1 - 4 + 9 - 16 = -(1 + 2 + 3 + 4)

Gu..."

2019-11-16T22:35:05Z

Created page with "{{FMM |title=Inferring Rule from Sequence |problem= <pre> 1 = 1 1 - 4 = -(1 + 2) 1 - 4 + 9 = 1 + 2 + 3 1 - 4 + 9 - 16 = -(1 + 2 + 3 + 4) </pre> Gu..."

New page

{{FMM
|title=Inferring Rule from Sequence
|problem=

<pre>
1 = 1
1 - 4 = -(1 + 2)
1 - 4 + 9 = 1 + 2 + 3
1 - 4 + 9 - 16 = -(1 + 2 + 3 + 4)
</pre>

Guess the general law suggested by these examples, express it in suitable mathematical notation, and prove it.

|solution=

The general law is:

1 - 4 + 9 - 16 + ... + (-1)^(n-1) n^2
=
(-1)^(n-1) (n(n+1))/2

<math>
\sum_{k=1}^n (-1)^{k-1} k^2 = (-1)^{n-1} \frac{(n(n+1))}{2}
</math>

This can be proved for the base case of n=1, in which case above equation reduces to 1=1

Then can show that if assume n is true, then n+1 holds

<math>
\sum_{k=1}^{n+1} (-1)^{k-1} k^2 = (-1)^{n} \frac{((n+1)(n+2))}{2}
</math>

If you split the left side into the sum from k=1 to n, (which can be reduced to a single term by using the identity above for the case of n, which you assume is true), and the last n+1 term written explicitly, you can do some algebraic manipulation to show it's equivalent to the right side.
}}