JEE Main Questions
Find n, if the ratio of the fifth term from the beginning to the fifth term
from the end in the expansion of (42+431)nis 6:1.
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The value ofr=0∑10(−1)r.410−r.30Cr.30−rC10−r is equal to
If the coefficients of three consecutive terms in the expansion of (1+x)n
are in the ratio 1:7:42, then find the value of n˙
Find the sum 2..10C0+222.10C1+323.10C2+424.10C3+....+11211.10C10.
The value of n.nC0+n+1.nC1+n+2.nC2+...+2n.nC2
is very large as compare to y,
prove that x+yxx−yx˙=1+2x2y2
The smallest integer larger than (3+2)6 is
Find the value of .20C0×.13C10−.20C1×.12C9+.20C2×.11C8−……+.20C10.
Prove that .nC0+5×.nC1+9×.nC2+….+(4n+1)×.nCn=(2m+1)2n.