Question

Popular Math Problem
Evaluate 100^2 - 99^2 + 98^2 - 97^2 + 96^2 - 95^2 + ... + 2^2 - 1^2 =?

Answer 1

Answer:

Value of 100²-99²+98²-97²+96²-95²+....+2²-1²=5050

Step-by-step explanation:

Given 100²-99²+98²-97²+96²-95²+....+2²-1²

= (100²-99²)+(98²-97²)+(96²-95²)+...+(2²-1²)

= (100+99)(100-99)+(98+97)(98-97)+(96+95)(96-95)+....+(2+1)(2-1)

/* By algebraic identity :

a²-b² = (a+b)(a-b) */

= (100+99)1+(98+97)1+(96+95)1+....+(2+1)1

= 100+99+98+97+96+95+....+2+1

$= \frac{100(100+1)}{2}$

$Sum \: of \: first \: n \: natural \: numbers \\= \frac{n(n+1)}{2}$

$= 50\times 101\\= 5050$

Therefore,

Value of 100²-99²+98²-97²+96²-95²+....+2²-1²=5050

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Answer 2

Answer:

5050

Step-by-step explanation:

⇒ 100² - 99² + 98² - 97² + 96² - 95² + ...2² - 1²

⇒ (100² - 99²) + (98² - 97²) + (96² - 95²) + ... (2² - 1²)

   Using a² - b² = (a + b)(a - b)

⇒ (199) + (195) + (191) + ... 3

This forms an arithematic series in which a = 199 and l = 3, and n = 100/2 = 50.

∴ S = (n/2) [1st term + last term]

      = (50/2) [199 + 3]

      = 5050

Hence, required sum is 5050.