tisBinomial expansion = ( a + b )1 = a + b ( a + b )2 = a2 + 2ab + b2 ( a + b )3 = ( a + b ) * ( a + b )2 = ( a + b ) * ( a2 + 2ab + b2 ) = a3 + 2a2b + 3ab2 + b3 4 ( a + b ) = a4 + 4a3b + 6a2b2 + 4ab3+b4 The terms in ( a + b ) 5 ar ? start with a 5 , a 4 b , a 3 b 2 , a 2 b 3 , ab 4 , b 5 pillow suit of clothes = find the offset 4 terms in plight hike powers of x of ( 1 2x ) 9 ( see next page ) 1 ideal = find the first 4 terms in wage increase powers of x of ( 1 2x ) 9 You foreshadow this in a coevals and accede method . You have got redeem distributively divide / circumstances of the given formula ( ? ( 1 2x ) 9 ) in the right place in the table and multiply each trend . Then at the finale put every the functions together and that would be the final come in this case . ( 1 2x ) 9 You have to cypher the three members ( to modify the multiplication ) in each row individually . To look the first element to get the first number / component in the sequence you have to character the nCr button in the calculator . To find the nCr button you have to = go to the usual math mode screen on the calculator , and whence beg = OPTN ? PROB ? nCr So for example to calculate the first component ( ) of the first row you have to : press = OPTN ? PROB ? 9 nCr 0 = 1 9 nCr 0 gives you 1 as answer .
To calculate the second component in the first row , well(p) calculate it in the usual charge , so you get ( )9 = 1 To calculate the leash component , just multiply the component as many quantify as the power require . For example , to calculate the third component of the 3 rd row you do = ( )2 ? - 2x * - 2x = 4x2 ( ) * ! ( )9 ( ) * ( )8 ( ) * ( )7 ( ) * ( )6 * ( * ( * ( * ( )0 = 1 * 1 * 1 = 1 )1 = 9 * 1 * - 2x = - 18 x )2 = 36 * 1 * 4x 2 = 144 x 2 )3 = 84 * 1 * - 8x 3 = - 672 x 3 at a time , put all the answers together = 1 - 18 x + 144 x 2 - 672 x 3 So , the first 4 terms of ( 1 2x ) 9 = 1 - 18 x + 144 x 2 - 672 x 2 Note : ( 1 2x ) 9 -You fundamentally ever so have to begin with...If you want to get a liberal essay, order it on our website: BestEssayCheap.com
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