Find the sum to n terms of the series: 5+11+1 | vedclass

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

Question

(D) Let the $n^{th}$ term be $a_n$. The series is $5, 11, 19, 29, 41, \ldots$The differences between consecutive terms are $6, 8, 10, 12, \ldots$,whic

Vedclass · Accepted Answer

$\frac{n(n+2)(n+4)}{3}$

Vedclass · Answer

(D) Let the $n^{th}$ term be $a_n$. The series is $5, 11, 19, 29, 41, \ldots$The differences between consecutive terms are $6, 8, 10, 12, \ldots$,which form an arithmetic progression.Thus,$a_n = An^2 + Bn + C$.For $n=1, A+B+C=5$.For $n=2, 4A+2B+C=11$.For $n=3, 9A+3B+C=19$.Subtracting the equations,we get $3A+B=6$ and $5A+B=8$. Solving these,$2A=2 \implies A=1$,$B=3$,and $C=1$.So,$a_n = n^2 + 3n + 1$.Now,$S_n = \sum_{k=1}^n (k^2 + 3k + 1) = \sum_{k=1}^n k^2 + 3\sum_{k=1}^n k + \sum_{k=1}^n 1$.$S_n = \frac{n(n+1)(2n+1)}{6} + \frac{3n(n+1)}{2} + n$.$S_n = \frac{n(2n^2 + 3n + 1) + 9n(n+1) + 6n}{6} = \frac{2n^3 + 3n^2 + n + 9n^2 + 9n + 6n}{6} = \frac{2n^3 + 12n^2 + 16n}{6} = \frac{n(n^2 + 6n + 8)}{3} = \frac{n(n+2)(n+4)}{3}$.

Find the sum to $n$ terms of the series: $5+11+19+29+41 + \ldots$

Similar Questions

The sum $1 + \frac{1}{2}(1^2+2^2) + \frac{1}{3}(1^2+2^2+3^2) + \dots$ up to $10$ terms is equal to:

If $\sum_{k=1}^n \left( \sum_{m=1}^k m^2 \right) = an^4 + bn^3 + cn^2 + dn + e$,then which of the following is true?

Write the first five terms of the sequence whose $n^{th}$ term is $a_{n} = n \frac{n^{2}+5}{4}$.

Show that $\frac{1 \times 2^{2}+2 \times 3^{2}+\ldots+n \times(n+1)^{2}}{1^{2} \times 2+2^{2} \times 3+\ldots+n^{2} \times(n+1)}=\frac{3 n+5}{3 n+1}$

The sum of the series $6 + 66 + 666 + \dots$ up to $n$ terms is

Vedclass Test Series

Exam Paper Generator

Online Exam Module