wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Let m be the smallest positive integer such that the coefficient of x2 in the expansion of (1+x)2+(1+x)3+...+(1+x)49+(1+mx)50 is (3n+1)51C3 for some positive interger n. Then the value of n is ___

Open in App
Solution

(1+x)2+(1+x)3+...+(1+x)49+(1+mx)50=(1+x)2[(1+x)48(1+x)1]+(1+mx)50=1x[(1+x)50(1+x)2)]+(1+mx)50coeff. of x2in the above expansion=coeff. of x3 in (1+x)50+Coeff. of x2in (1+mx)5050C3+50C2m2(3n+1)51C3=50C3+50C2m2(3n+1)=50C351C3+50C251C3m23n+1=1617+117m2n=m2151
Least positive integer m for which n is an interger is m=16 and then n=5

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Visualising the Coefficients
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon