The sum of the series 1+3+5^2+7+9^2+ldots up | vedclass

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(B) The series is $1, 3, 5^2, 7, 9^2, 11, 13^2, \ldots$ up to $40$ terms.We can split this into two series:Series $1$: $1, 5^2, 9^2, \ldots$ ($20$ ter

Vedclass · Accepted Answer

$41880$

Vedclass · Answer

(B) The series is $1, 3, 5^2, 7, 9^2, 11, 13^2, \ldots$ up to $40$ terms.We can split this into two series:Series $1$: $1, 5^2, 9^2, \ldots$ ($20$ terms) where the $r$-th term is $(4r-3)^2$.Series $2$: $3, 7, 11, \ldots$ ($20$ terms) where the $r$-th term is $(4r-1)$.Sum $= \sum_{r=1}^{20} (4r-3)^2 + \sum_{r=1}^{20} (4r-1)$$= \sum_{r=1}^{20} (16r^2 - 24r + 9 + 4r - 1)$$= \sum_{r=1}^{20} (16r^2 - 20r + 8)$$= 16 \sum_{r=1}^{20} r^2 - 20 \sum_{r=1}^{20} r + 8 \sum_{r=1}^{20} 1$$= 16 \left( \frac{20 	imes 21 	imes 41}{6} ight) - 20 \left( \frac{20 	imes 21}{2} ight) + 8(20)$$= 16(2870) - 20(210) + 160$$= 45920 - 4200 + 160 = 41880$.

The sum of the series $1+3+5^2+7+9^2+\ldots$ up to $40$ terms is equal to

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