Difference between revisions of "Talk:Introduction-TADM2E"
From Algorithm Wiki
(Solution for 1-10) |
(Solution for TADM2E 1.10 and 1.12) |
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| − | = TADM2E 1.10 = | + | As some even number problem's solution has already been provided so please find solution for following |
| + | ---- | ||
| + | |||
| + | == TADM2E 1.10 == | ||
First we'll verify the base case for <math>n = 1</math> : | First we'll verify the base case for <math>n = 1</math> : | ||
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: <math>\sum_{i=1}^n i = \sum_{i=1}^{n-1} i + n = \frac{(n-1)n}{2} + n = \frac{n^2 - n + 2n}{2} = \frac{n^2 + n}{2} = \frac{n(n+1)}{2}</math> | : <math>\sum_{i=1}^n i = \sum_{i=1}^{n-1} i + n = \frac{(n-1)n}{2} + n = \frac{n^2 - n + 2n}{2} = \frac{n^2 + n}{2} = \frac{n(n+1)}{2}</math> | ||
| + | |||
| + | ---- | ||
| + | |||
| + | == TADM2E 1.12 == | ||
| + | |||
| + | First we'll verify the base case for <math> n = 1</math> : | ||
| + | |||
| + | :<math>\sum_{i=1}^1 i^3 = 1^3 = \frac{(1)^2(1+1)^2}{4} = 1</math> | ||
| + | |||
| + | Now, we'll assume given statement is true up to <math>n - 1</math>.<br> | ||
| + | So, following statement is true. | ||
| + | :<math>\sum_{i=1}^{n-1} i^3 = \frac{(n-1)^2(n-1+1)^2}{4} = \frac{(n-1)^2n^2}{4}</math> | ||
| + | |||
| + | To prove for the general case <math>n</math>, | ||
| + | |||
| + | :<math> | ||
| + | \begin{align} | ||
| + | \sum_{i=1}^{n} i^3 & = \sum_{i=1}^{n-1} i^3 + n^3 \\ | ||
| + | & = \frac{(n-1)^2n^2}{4} + n^3 \\ | ||
| + | & = \frac{(n^2 - 2n + 1)n^2 + 4n^3}{4} \\ | ||
| + | & = \frac{n^4 - 2n^3 + n^2 + 4n^3}{4} \\ | ||
| + | & = \frac{n^4 + 2n^3 + n^2}{4} \\ | ||
| + | & = \frac{n^2(n^2 + 2n + 1)}{4} \\ | ||
| + | \sum_{i=1}^{n} i^3 & = \frac{n^2(n + 1)^2}{4} \\ | ||
| + | \end{align} | ||
| + | </math> | ||
Revision as of 08:18, 9 October 2015
As some even number problem's solution has already been provided so please find solution for following
TADM2E 1.10
First we'll verify the base case for $ n = 1 $ :
- $ \sum_{i=1}^1 i = \frac{1(1+1)}{2} = 1 $
Now, we'll assume given statement is true up to $ n - 1 $.
So,
- $ \sum_{i=1}^{n-1} i = \frac{(n-1)(n-1+1)}{2} = \frac{(n-1)n}{2} $
To prove for the general case $ n $,
- $ \sum_{i=1}^n i = \sum_{i=1}^{n-1} i + n = \frac{(n-1)n}{2} + n = \frac{n^2 - n + 2n}{2} = \frac{n^2 + n}{2} = \frac{n(n+1)}{2} $
TADM2E 1.12
First we'll verify the base case for $ n = 1 $ :
- $ \sum_{i=1}^1 i^3 = 1^3 = \frac{(1)^2(1+1)^2}{4} = 1 $
Now, we'll assume given statement is true up to $ n - 1 $.
So, following statement is true.
- $ \sum_{i=1}^{n-1} i^3 = \frac{(n-1)^2(n-1+1)^2}{4} = \frac{(n-1)^2n^2}{4} $
To prove for the general case $ n $,
- $ \begin{align} \sum_{i=1}^{n} i^3 & = \sum_{i=1}^{n-1} i^3 + n^3 \\ & = \frac{(n-1)^2n^2}{4} + n^3 \\ & = \frac{(n^2 - 2n + 1)n^2 + 4n^3}{4} \\ & = \frac{n^4 - 2n^3 + n^2 + 4n^3}{4} \\ & = \frac{n^4 + 2n^3 + n^2}{4} \\ & = \frac{n^2(n^2 + 2n + 1)}{4} \\ \sum_{i=1}^{n} i^3 & = \frac{n^2(n + 1)^2}{4} \\ \end{align} $
