Using r+1(rn)=n+11(r+1n+1):
r=50∑100r+1(r100)=1011r=50∑100(r+1101)=1011k=51∑101(k101).
Since 101 is odd, by symmetry k=51∑101(k101)=k=0∑50(k101)=22101=2100.
Required value =1012100.
JEE Main 2026 — Mathematics Algebra
The value of 51100C50+52100C51+….+101100C100 is :
Held on 23 Jan 2026 · Verified 6 Jul 2026.
1002101
1002100
1012101
1012100
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