Answered: or every integer n ∈ ℕ, it follows that…

Linear Algebra: A Modern Introduction

4th Edition

ISBN:9781285463247

Author:David Poole

Publisher:David Poole

Chapter7: Distance And Approximation

Section7.2: Norms And Distance Functions

Problem 14EQ

See similar textbooks

Related questions

Question

for every integer n ∈ ℕ, it follows that 1^2, 2^2, 3^2, 4^2 + ...+ n^2 = (n(n+1)(2n+1)/)/6. Prove using S_x

Expert Solution

Step 1

Claim: $1^{2} + 2^{2} + 3^{2} + \dots + n^{2} = \frac{n (n + 1) (2 n + 1)}{6}$

Proof: Now ${(k - 1)}^{3} = k^{3} - 3 k^{2} + 3 k - 1$

Rearranging the above equation. we get,

$k^{3} - {(k - 1)}^{3} = 3 k^{2} - 3 k + 1$

Taking summation on both sides from $k = 1$ to $n$ , gives,

$\begin{array}{rcl} \sum_{k = 1}^{n} [k^{3} - {(k - 1)}^{3}] & = & \sum_{k = 1}^{n} 3 k^{2} - \sum_{k = 1}^{n} 3 k + \sum_{k = 1}^{n} 1 \\ 1^{3} - 0^{3} + 2^{3} - 1^{3} + 3^{3} - 2^{3} + 4^{3} - 3^{3} + \dots + {(n - 1)}^{3} - {(n - 2)}^{3} + n^{3} - {(n - 1)}^{3} & = & 3 \sum_{k = 1}^{n} k^{2} - 3 \sum_{k = 1}^{n} k + n \\ n^{3} & = & 3 \sum_{k = 1}^{n} k^{2} - 3 \frac{n (n + 1)}{2} + n \\ 3 \sum_{k = 1}^{n} k^{2} & = & n^{3} + 3 \frac{n (n + 1)}{2} - n \\ \sum_{k = 1}^{n} k^{2} & = & \frac{n^{3}}{3} + \frac{n (n + 1)}{2} - \frac{n}{3} \\ \sum_{k = 1}^{n} k^{2} & = & \frac{n^{3}}{3} + \frac{n^{2}}{2} + \frac{n}{2} - \frac{n}{3} \\ \sum_{k = 1}^{n} k^{2} & = & \frac{n^{3}}{3} + \frac{n^{2}}{2} + \frac{n}{6} \\ \sum_{k = 1}^{n} k^{2} & = & \frac{2 n^{3} + 3 n^{2} + n}{6} \\ \sum_{k = 1}^{n} k^{2} & = & \frac{2 n^{3} + n^{2} + 2 n^{2} + n}{6} \\ \sum_{k = 1}^{n} k^{2} & = & \frac{n^{2} (2 n + 1) + n (2 n + 1)}{6} \\ \sum_{k = 1}^{n} k^{2} & = & \frac{(n^{2} + n) (2 n + 1)}{6} \\ \sum_{k = 1}^{n} k^{2} & = & \frac{n (n + 1) (2 n + 1)}{6} \end{array}$

Trending now

This is a popular solution!

Step by step

Solved in 2 steps

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

$Complexity$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, advanced-math and related others by exploring similar questions and additional content below.

Similar questions