Parth: Another solution for TADM2E 1.11

2015-10-09T06:53:02Z

Another solution for TADM2E 1.11

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Instead of assuming that summation is true for <math>n</math>, we can choose to assume it for <math>n-1</math> that way it'll be easy to understand and more intuitive.

Anyway I'm writing the solution for same,

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First we'll verify the base case for <math>n = 1</math> :

:<math>\sum_{i=1}^1 i^2 = 1^2 = \frac{1(1+1)(2+1)}{6} = 1</math>

Now, we'll assume given statement is true up to <math>n - 1</math>.<br>
So, following is true.

:<math>
\begin{align}
\sum_{i=1}^{n-1} i^2 & = \frac{(n-1)(n-1+1)(2(n-1)+1)}{6} \\
& = \frac{n(n-1)(2n-1)}{6} \\
\end{align}
</math>

To prove the general case <math>n</math>,

:<math>
\begin{align}
\sum_{i=1}^n i^2 & = \sum_{i=1}^{n-1} i^2 + n^2 \\
& = \frac{n(n-1)(2n-1)}{6} + n^2 \\
& = \frac{(n^2-n)(2n-1) + 6n^2}{6} \\
& = \frac{2n^3 + 3n^2 + n}{6} \\
& = \frac{n(2n^2 + 3n + 1)}{6} \\
& = \frac{n(2n^2 + 2n + n + 1)}{6} \\
& = \frac{n(2n(n + 1) + n + 1)}{6} \\
\sum_{i=1}^n i^2 & = \frac{n(n + 1)(2n + 1)}{6} \\
\end{align}
</math>

Talk:TADM2E 1.11 - Revision history

Parth: Another solution for TADM2E 1.11