If 2.5+5.9+8.13+11.17+ldots to n terms =an^3+ | vedclass

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(D) The $n$-th term of the series is $T_n = (3n-1)(4n+1) = 12n^2 - n - 1$.Sum of $n$ terms $S_n = \sum_{k=1}^{n} (12k^2 - k - 1) = 12 \sum_{k=1}^{n} k

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-$1$

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(D) The $n$-th term of the series is $T_n = (3n-1)(4n+1) = 12n^2 - n - 1$.Sum of $n$ terms $S_n = \sum_{k=1}^{n} (12k^2 - k - 1) = 12 \sum_{k=1}^{n} k^2 - \sum_{k=1}^{n} k - \sum_{k=1}^{n} 1$.Using standard summation formulas:$S_n = 12 \frac{n(n+1)(2n+1)}{6} - \frac{n(n+1)}{2} - n$.$S_n = 2n(2n^2+3n+1) - \frac{n^2+n}{2} - n = 4n^3 + 6n^2 + 2n - 0.5n^2 - 0.5n - n$.$S_n = 4n^3 + 5.5n^2 + 0.5n$.Comparing with $an^3+bn^2+cn+d$,we get $a=4, b=5.5, c=0.5, d=0$.Then $a-b+c-d = 4 - 5.5 + 0.5 - 0 = -1$.

If $2.5+5.9+8.13+11.17+\ldots$ to $n$ terms $=an^3+bn^2+cn+d$,then $a-b+c-d=$

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