Chapter 4: Q. 15 (page 325)

Verify that ${\sum k}_{k = 1}^{n} is equal to \frac{n (n + 1)}{2}$ for the cases $(a) n = 2, (b) n = 8, (c) n = 9$

Short Answer

Expert verified

$Part (a) F o r n = 2, {\sum k}_{k = 1}^{n} = \frac{n (n + 1)}{2} Part (b) F o r n = 8, {\sum k}_{k = 1}^{n} = \frac{n (n + 1)}{2} Part (c) F o r n = 9, {\sum k}_{k = 1}^{n} = \frac{n (n + 1)}{2}$

Step by step solution

Part(a) Step 1. Given Information

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

Part(a) Step 2. Calculation

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

${\sum k}_{k = 1}^{2} = 1 + 2 = 3 a n d \frac{2 (2 + 1)}{2} = \frac{2 (3)}{2} = 3$

Thus, it is verified.

Part(b) Step 1. Calculation

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

{\sum k}_{k = 1}^{8} = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36 a n d \frac{8 (8 + 1)}{2} = \frac{8 (9)}{2} = 36

Thus, it is verified.

Part(c) Step 1. Calculation

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

${\sum k}_{k = 1}^{9} = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45 a n d \frac{9 (9 + 1)}{2} = \frac{9 (10)}{2} = 45$

Thus, it is verified.

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Verify that ${\sum k}_{k = 1}^{n} is equal to \frac{n (n + 1)}{2}$ for the cases $(a) n = 2, (b) n = 8, (c) n = 9$

Short Answer

Step by step solution

Part(a) Step 1. Given Information

Part(a) Step 2. Calculation

Part(b) Step 1. Calculation

Part(c) Step 1. Calculation

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Verify that∑kk=1nisequalton(n+1)2for the cases(a)n=2,(b)n=8,(c)n=9

Short Answer

Step by step solution

Part(a) Step 1. Given Information

Part(a) Step 2. Calculation

Part(b) Step 1. Calculation

Part(c) Step 1. Calculation

One App. One Place for Learning.

Most popular questions from this chapter

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Verify that ${\sum k}_{k = 1}^{n} is equal to \frac{n (n + 1)}{2}$ for the cases $(a) n = 2, (b) n = 8, (c) n = 9$