Question
श्रेणी $5+11+19+29+41+\ldots$ के प्रथम $20$ पदों का योग है
Let $T _{ r }=a r^2+ br + c$
$T _1= a + b + c =5$
$T _2=4 a +2 b + c =11$
$T _3=9 a +3 b + c =19$
$a =1, b =3, c =1$
Hence $S _{20}=\sum_{ r =1}^{20} r ^2+3 \sum_{ r =1}^{20} r +\sum_{ r =1}^{20} 1=3520$
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