Question

OBLEMS FOR PRACT
n -
Evaluate : E(8r – 7)
r=1​

Answer 1

EXPLANATION.

Evaluate :

$\sf \displaystyle \sum_{r = 1}^{n} (8r - 7)$

Put the value of r = 1, 2, 3, . . . . . upto n terms, we get.

$\sf \displaystyle [8(1) - 7] + [8(2) - 7] + [8(3) - 7] + [8 (4) - 7] + . . . . . + [8(n) - 7]$

$\sf \displaystyle (8 - 7) + (16 - 7) + (24 - 7) + (32 - 7) + . . . . . + (8n - 7)$

$\sf \displaystyle 1 + 9 + 17 + 25 + . . . . . + (8n - 7)$

We get Arithmetic series.

As we know that,

Formula of :

Sum of first n terms in an ap.

⇒ Sₙ = n/2[2a + (n - 1)d].

Using this formula in this question, we get.

First term : a = 1.

Common difference : d = b - a : 9 - 1 = 8.

$\sf \displaystyle S_{n} = \frac{n}{2} \bigg[2(1) + (n - 1)(8) \bigg]$

$\sf \displaystyle S_{n} = \frac{n}{2} (2 + 8n - 8)$

$\sf \displaystyle S_{n} = \frac{n}{2} (8n - 6)$

$\sf \displaystyle S_{n} = n( 4n - 3)$

$\sf \displaystyle S_{n} = 4n^{2} - 3n$

$\sf \displaystyle \boxed{ \sum_{r = 1}^{n} (8r - 7) = 4n^{2} - 3n}$

Answer 2

By evaluate the summation expression that is $\sum_{r=1}^n (8r-7)$ the solved solution is n(4n-3).

Given that,

We have to evaluate the summation expression that is $\sum_{r=1}^n (8r-7)$

We know that,

What is summation?

The sum of the series is calculated using the summation formulas. There are many different kinds of sequences, including arithmetic and geometric sequences, and consequently, there are many different kinds of summing formulas for those different kinds of sequences.

So,

$\sum_{r=1}^nr = \frac{n(n+1)}{2}$

$\sum_{r=1}^nk = k_n$             [where k = constant]

$\sum_{r=1}^n (8r-7) = \sum_{r=1}^n 8r - \sum_{r=1}^n 7$

$\sum_{r=1}^n (8r-7) = 8\sum_{r=1}^n r - \sum_{r=1}^n 7$

$\sum_{r=1}^n (8r-7) = 8\frac{n(n+1)}{2} - 7n$

= 4n(n+1) - 7n

=4n² + 4n -7n

= 4n² - 3n

= n(4n-3)

Therefore, By evaluate the summation expression that is $\sum_{r=1}^n (8r-7)$ the solved solution is n(4n-3).

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