Question

the value of (2^-4)^2 × 2-^5/2-^6​

Answer 1

Answer:

(n+1)^2 - n^2 = n^2 + 2n + 1 - n^2 = 2n + 1 = n + (n+1)

This serious can be rearranged as:

(2^2 - 1^2) + (4^2 - 3^2) + ….. (100^2 - 99^2)

= (1 + 2) + (3+4) + …… +(99+100)

Which is the sum of first 100 natural numbers, which is calculated as n*(n+1)/2, which is 100*101/2 = 5050.