JEE Main20244 Apr 2024Evening ShiftMathematicsSequences and SeriesActual
The value of 1 2^2+2 3^2+ +100 (101)^2 1^2 2+2^2 3+ .+100^2 101 is
Options
- A32 31
- B31 30
- C306 305
- D305 301
Correct answer
D. 305 301
Step-by-step solution
aligned & 1 2^2+2 3^2+ +100 (101)^2 1^2 2+2^2 3+ +100^2 101 = _ r=1 ¹⁰⁰ r(r+1)^2 _ r=1 ¹⁰⁰ r^2(r+1) & = _ r=1 ¹⁰⁰ (r^3+2 r^2+r ) _ r=1 ¹⁰⁰ (r^3+r^2 ) = ( n(n+1)^2 2 )+ 2 n(n+1)(2 n+1) 6 + n(n+1) 2 ( n(n+1) 2 )^2+ n(n+1)(2 n+1) 6 & = n(n+1) 2 [ n(n+1) 2 + 2 3 (2 n+1)+1 ] n(n+1) 2 [ n(n+1) 2 + (2 n+1) 3 ] ; Put n=100 & = 100(101) 2 + 2 3 (201)+1 100 101 2 + 201 3 = 5185 5117 = 305 301 & aligned