If sum_{i = 1}^n i = frac{n(n + 1)}{2},then s | vedclass

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

Question

(A) Given the summation $\sum_{i = 1}^n (3i - 2)$.Using the linearity property of summation,we can write:$\sum_{i = 1}^n (3i - 2) = 3 \sum_{i = 1}^n i

Vedclass · Accepted Answer

$\frac{n(3n - 1)}{2}$

Vedclass · Answer

(A) Given the summation $\sum_{i = 1}^n (3i - 2)$.Using the linearity property of summation,we can write:$\sum_{i = 1}^n (3i - 2) = 3 \sum_{i = 1}^n i - \sum_{i = 1}^n 2$.Since $\sum_{i = 1}^n i = \frac{n(n + 1)}{2}$ and $\sum_{i = 1}^n 2 = 2n$,we substitute these values:$= 3 \left[ \frac{n(n + 1)}{2} \right] - 2n$.$= \frac{3n^2 + 3n}{2} - 2n$.$= \frac{3n^2 + 3n - 4n}{2}$.$= \frac{3n^2 - n}{2} = \frac{n(3n - 1)}{2}$.

If $\sum_{i = 1}^n i = \frac{n(n + 1)}{2}$,then $\sum_{i = 1}^n (3i - 2) = $

Similar Questions

Let $b_1, b_2, \dots, b_n$ be a geometric sequence such that $b_1 + b_2 = 1$ and $\sum_{k=1}^{\infty} b_k = 2$. Given that $b_2 < 0$,then the value of $b_1$ is:

The sum to $n$ terms of the series $1 \cdot 3^2 + 2 \cdot 5^2 + 3 \cdot 7^2 + \dots$ is

If $\log _{10} 2, \log _{10} (2^x - 1), \log _{10} (2^x + 3)$ are in $A.P.,$ then :-

If the $p^{th}$ term of an $A.P.$ is $\frac{1}{q}$ and the $q^{th}$ term is $\frac{1}{p}$,then the sum of its $pq$ terms will be:

Suppose $a, b, c$ are in $A.P.$ and $a^2, b^2, c^2$ are in $G.P.$ If $a < b < c$ and $a + b + c = \frac{3}{2}$,then the value of $a$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module