If alpha in R, n in N and n+2(n-1)+3(n-2)+ldo | vedclass

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

Question

(D) The general term of the series is $T_r = r(n-r+1) = nr - r^2 + r$. Summing from $r=1$ to $n$: $S_n = \sum_{r=1}^n (nr - r^2 + r) = (n+1) \sum_{r=1

Vedclass · Accepted Answer

$\frac{1}{6}$

Vedclass · Answer

(D) The general term of the series is $T_r = r(n-r+1) = nr - r^2 + r$. Summing from $r=1$ to $n$: $S_n = \sum_{r=1}^n (nr - r^2 + r) = (n+1) \sum_{r=1}^n r - \sum_{r=1}^n r^2$. Using standard summation formulas: $S_n = (n+1) \frac{n(n+1)}{2} - \frac{n(n+1)(2n+1)}{6}$. Factoring out $\frac{n(n+1)}{2}$: $S_n = \frac{n(n+1)}{2} [ (n+1) - \frac{2n+1}{3} ] = \frac{n(n+1)}{2} [ \frac{3n+3-2n-1}{3} ] = \frac{n(n+1)(n+2)}{6}$. Comparing this with $\alpha n(n+1)(n+2)$,we get $\alpha = \frac{1}{6}$.

If $\alpha \in R, n \in N$ and $n+2(n-1)+3(n-2)+\ldots+(n-1)2+n.1 = \alpha n(n+1)(n+2)$,then $\alpha =$

Similar Questions

The sum of an infinite number of terms in a $G.P.$ is $20$ and the sum of their squares is $100$. The common ratio of the $G.P.$ is:

If $S_n = 2 + 4 + 7 + 11 + \dots + n$ terms,then $t_n = \dots$

The sum of the series $1 \cdot 3 \cdot 5 + 2 \cdot 5 \cdot 8 + 3 \cdot 7 \cdot 11 + \dots$ up to $n$ terms is:

The expression for $a_n$ which satisfies $a_0=0, a_1=1$ and $a_n=a_{n-1}+a_{n-2}, \forall n \in N -\{0,1\}$ is:

If the $n^{th}$ term of a sequence is $n(n + 1)$,then what is the sum of its $n$ terms?

Vedclass Test Series

Exam Paper Generator

Online Exam Module