IIT JEE , AIEEE Forum - Mathematics , Algebra

shrithi

given x,y,n r positive integers and...y²=_{[r=1 ]}

^{[ r=n]} r! and...x²=_{[ r=1]}

^{[ n]} (r!)³...then x²-y²=------------------

Decoder

nice question..

u get x²-y² =

r! ( r! +1)(r!-1)..... solving r!³ - r! ..

here lies the tricky part..

now divide and multiply the expression by { (r! + 2) - (r!-2) } ...

it is nothing but 4..

so expression gets..

x²-y² = 1/4

r! ( r! +1)(r!-1)(r!+2) - r! ( r! +1)(r!-1)(r!-2)

observe carefully by putting values..u will observe tht terms r cancelling out..

at last u r left with 1/4 { (n!-1)(n!)(n!+1)(n!+2) - 1 }..

shrithi

how to solve the last step coz the options r as follows

a)216 b)220 c)192 d)236

shrithi

can u plz explain??

Decoder

u shud noe sumthing abt n...otherwise..waise bhi iska koi integral answer nahi ho sakta..

shrithi

acha...r u sure? ..kyunki it was given tht way in my question paper today