Question

2020^2-2019^2+2018^2-2017^2+.....+4^2-3^2+2^2-1^2 =x then find x​

Answer 1

Answer:

tried for that answer for the given problem is given

Answer 2

SOLUTION

GIVEN

$\sf {2020}^{2} - {2019}^{2} + {2018}^{2} - {2017}^{2} + ... + {4}^{2} - {3}^{2} + {2}^{2} - {1}^{2} = x$

TO DETERMINE

The value of x

EVALUATION

$\sf {2020}^{2} - {2019}^{2} + {2018}^{2} - {2017}^{2} + ... + {4}^{2} - {3}^{2} + {2}^{2} - {1}^{2} = x$

$\sf \implies x = {2020}^{2} - {2019}^{2} + {2018}^{2} - {2017}^{2} + ... + {4}^{2} - {3}^{2} + {2}^{2} - {1}^{2}$

Number of terms = 2020

We group them taking 2 terms in one bracket

$\sf \implies x =( {2020}^{2} - {2019}^{2}) + ( {2018}^{2} - {2017}^{2} ) + ... + ({4}^{2} - {3}^{2}) +( {2}^{2} - {1}^{2} )$

Number of brackets = 1010

$\sf \implies x =( 2020 + 2019)( 2020 - 2019) + (2018 + 2017)(2018 - 2017) + ... + (4 + 3)(4 - 3) +(2 + 1)(2 - 1)$

Now simplify the above

$\sf \implies x =4039 + 4035 + ... + 7 +3$

This is an arithmetic progression

First term = a = 4039

Common Difference = d = - 4

Number of terms = 1010

$\displaystyle \sf{ x= \frac{n}{2} \bigg[ 2a + (n - 1)d\bigg] }$

$\displaystyle \sf{ \implies x= \frac{1010}{2} \bigg[ (2 \times 4039) - 4 (1010 - 1)\bigg] }$

$\displaystyle \sf{ \implies x= 505 \bigg[8078 - 4 \times 1009\bigg] }$

$\displaystyle \sf{ \implies x= 505 \bigg[8078 - 4036\bigg] }$

$\displaystyle \sf{ \implies x= 505 \times 4042 }$

$\displaystyle \sf{ \implies x= 2041210 }$

FINAL ANSWER

Hence the required value of x = 2041210

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