The sum 1+3+11+25+45+71+ldots up to 20 terms | vedclass

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(A) Let the given series be $S_{20} = 1+3+11+25+45+71+\ldots+T_{20}$.The differences between consecutive terms are $2, 8, 14, 20, 26, \ldots$,which fo

Vedclass · Accepted Answer

$7240$

Vedclass · Answer

(A) Let the given series be $S_{20} = 1+3+11+25+45+71+\ldots+T_{20}$.The differences between consecutive terms are $2, 8, 14, 20, 26, \ldots$,which form an Arithmetic Progression ($A$.$P$.) with a common difference of $6$.Since the first-order differences are in $A$.$P$.,the general term $T_n$ is a quadratic expression of the form $T_n = an^2 + bn + c$.Using the first three terms:$T_1 = a + b + c = 1$$T_2 = 4a + 2b + c = 3$$T_3 = 9a + 3b + c = 11$Subtracting the equations:$(T_2 - T_1) \implies 3a + b = 2$$(T_3 - T_2) \implies 5a + b = 8$Subtracting these results: $2a = 6 \implies a = 3$.Substituting $a=3$ into $3a+b=2$,we get $9+b=2 \implies b = -7$.Substituting $a=3, b=-7$ into $a+b+c=1$,we get $3-7+c=1 \implies c = 5$.Thus,the general term is $T_n = 3n^2 - 7n + 5$.The sum of $20$ terms is $\sum_{n=1}^{20} (3n^2 - 7n + 5) = 3 \sum_{n=1}^{20} n^2 - 7 \sum_{n=1}^{20} n + \sum_{n=1}^{20} 5$.Using standard summation formulas:$= 3 \left( \frac{20(21)(41)}{6} \right) - 7 \left( \frac{20(21)}{2} \right) + 5(20)$$= 3(2870) - 7(210) + 100$$= 8610 - 1470 + 100 = 7240$.

The sum $1+3+11+25+45+71+\ldots$ up to $20$ terms is equal to:

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