Question

100^2-99^2+98^2-97^2_ _ _ _ _ _+2^2-1^2. ➡️ ➡️ ➡️ ➡️ please don't spam ​

Answer 1

⭐Solution⭐

➡given series

100²-99²+98²-97²+96²-95² + ...........+4²-3²+2²-1²

We know,

(x²-y²)=(x+y)(x-y).

so, can be write series

(100²-99²)+(98²-97²)+(96²-95²)+............+(4²-3²)+(2²-1²)

=>(100+99)(100-99)+(98+97)(98-97)+(96-95)(96+95)+..............(4+3)(4-3)+(2-1)(2+1)

=>(100+99)(1)+(98+97)(1)+......(4+3)(1)+(2+1)(1)

=>100+99+98+97+......+3+2+1

or,

=>1+2+3+........+98+99+100

This became an AP series ..where first term a =1, common difference d = 1.

✔Apply the formula to find sum of term of an AP

Sum= n/2[ 2a+(n-1)×d]

Sum = 100/2[ 2×1+(100-1)×1]

= 50[2+(99)]

= 50×101

= 5050

⏩Thus:-

Sum of given series will be

= 5050.

➡Hopes its help's u

⏩Please Marks Brainliest .