Question

find the sum of (40^2-39^2)+(38^2-37^2)+(36^2-35^2)​

Answer 1

➡️ Question :-

To find the sum of

$( {40}^{2} - {39}^{2} ) + ( {38}^{2} - {37}^{2} ) + ( {36}^{2} - {35}^{2})$

$\huge{\mathfrak{\underline{\red{Solution!}}}}$

By using the mathematical property,

${x}^{2} - {y}^{2} = (x - y)(x + y)$

where x and y are any of the numerical values.

$= ( {40}^{2} - {39}^{2} ) + ( {38}^{2} - {37}^{2} ) + ( {36}^{2} - {35}^{2})$

$= ( 40 - 39 )(40 + 39) + ( 38- 37) (38 + 37)+ ( 36- 35) (36 + 35)$

$= ( 1 )(79) + ( 1) (75)+ ( 1) ( 71)$

$= ( 1 \times 79) + ( 1 \times 75)+ ( 1 \times 71)$

$= 79 + 75+ 71$

$= \green{\boxed{\underline{225 }}} \: ans.$

Hope It Helps♡

Answer 2

➡️ Question :-

To find the sum of

answer is 225