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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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:

Similar Questions

The sum of the $1^{st} n$ terms of the series $\frac{1^{2}}{1} + \frac{1^{2}+2^{2}}{1+2} + \frac{1^{2}+2^{2}+3^{2}}{1+2+3} + \ldots$ is:

The numbers $a_n$ are defined by $a_0=1$ and $a_{n+1}=3n^2+n+a_n$ for $n \geq 0$. Then $a_n$ is equal to:

The product $(32)(32)^{1/6}(32)^{1/36} \dots \infty$ is

If $1+\frac{\cos \theta}{2}+\frac{\cos 2 \theta}{4}+\frac{\cos 3 \theta}{8}+\ldots = \frac{a-2 \cos \theta}{5+b \cos \theta}$ for some $a, b \in R$,then $(a-b)^2=$

If the $n^{th}$ term of a series is $3 + n(n - 1)$,then the sum of $n$ terms of the series is

Vedclass Test Series

Exam Paper Generator

Online Exam Module