Chapter 4: Q. 16 (page 325)

Verify that $\sum_{k = 1}^{n} k^{2} is equal to \frac{n (n + 1) (2 n + 1)}{6}$ for the cases $(a) n = 1, (b) n = 5, (c) n = 10$

Short Answer

Expert verified

$Part (a) F o r n = 1, \sum_{k = 1}^{n} k^{2} = \frac{n (n + 1) (2 n + 1)}{6} Part (b) F o r n = 5, \sum_{k = 1}^{n} k^{2} = \frac{n (n + 1) (2 n + 1)}{6} Part (c) F o r n = 10, \sum_{k = 1}^{n} k^{2} = \frac{n (n + 1) (2 n + 1)}{6}$

Step by step solution

Part(a) Step 1. Given Information

The given sum is $\sum_{k = 1}^{n} k^{2}$

Part(a) Step 2. Calculation

The left-hand side sum and the right-hand side sum is as follows,

$\sum_{k = 1}^{1} k^{2} = 1 A n d \frac{n (n + 1) (2 n + 1)}{6} = \frac{1 (1 + 1) (2 (1) + 1)}{6} = \frac{6}{6} = 1$

Thus, it is verified.

Part(c) Step 1. Calculation

The left-hand side sum and the right-hand side sum are as follows,

$\sum_{k = 1}^{5} k^{2} = 1^{2} + 2^{2} + 3^{2} + 4^{2} + 5^{2} = 55 A n d \frac{n (n + 1) (2 n + 1)}{6} = \frac{5 (5 + 1) (2 (5) + 1)}{6} = \frac{5 \times 6 \times 11}{6} = 55$

Thus, it is verified.

Part(c) Step 1. Calculation

The left-hand side sum and the right-hand side sum are as follows,

$\sum_{k = 1}^{10} k^{2} = 1^{2} + 2^{2} + 3^{2} + 4^{2} + 5^{2} + 6^{2} + 7^{2} + 8^{2} + 9^{2} + 10^{2} = 385 A n d \frac{n (n + 1) (2 n + 1)}{6} = \frac{10 (10 + 1) (2 (10) + 1)}{6} = \frac{10 \times 11 \times 21}{6} = 385$

Thus, it is verified.

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Verify that $\sum_{k = 1}^{n} k^{2} is equal to \frac{n (n + 1) (2 n + 1)}{6}$ for the cases $(a) n = 1, (b) n = 5, (c) n = 10$

Short Answer

Step by step solution

Part(a) Step 1. Given Information

Part(a) Step 2. Calculation

Part(c) Step 1. Calculation

Part(c) Step 1. Calculation

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Verify that $\sum_{k = 1}^{n} k^{2} is equal to \frac{n (n + 1) (2 n + 1)}{6}$ for the cases $(a) n = 1, (b) n = 5, (c) n = 10$