If the coefficient of x1274 in the expansion of (x+1)(x−2)2(x+3)3(x−4)4⋯⋯(x+49)49(x−50)50 is −k, then the value of k is
Open in App
Solution
Degree of expansion is 1+2+3+⋯+50=50×512=1275
Coefficient of x1274 =1−2C1(2)+3C2(3)−4C3(4)+⋯=12−22+32−42+⋯+492−502=−3−7−…−99=−(3+7+…+99)
It is an A.P. where a=3,d=4 99=3+4(n−1)⇒n=25
Therefore, the required coefficient =−252[3+99]=25×51=−1275∴k=1275