Sn=k=1∑nK2+5k+64=k=1∑n(K+2)(K+3)4=4K=1∑n(K+21−K+31) =4[31−41]=4[41−51]=4[n+21−n+31]Sn=4[31−n+31] S2025=4[31−20281]S2025=4[2028675]507S2025=675
JEE Main 2025 — Mathematics Algebra
For positive integers n, if 4an=(n2+5n+6) and Sn=k=1∑n(ak1), then the value of 507S2025 is :
Held on 28 Jan 2025 · Verified 6 Jul 2026.
540
675
1350
135
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