Question

3 + 32 + 33 + ... + 38 =?​

Answer 1

$\underline{\textbf{Given:}}$

$\mathsf{31+32+33+\;.\;.\;.\;.\;.+38}$

$\underline{\textbf{To find:}}$

$\textsf{Value of the sum}$

$\mathsf{31+32+33+\;.\;.\;.\;.\;.+38}$

$\underline{\textbf{Solution:}}$

$\underline{\textbf{Sum of first 'n' natural numbers:}}$

$\boxed{\mathsf{1+2+3+\;.\;.\;.\;.\;.\;.+n=\dfrac{n(n+1)}{2}}}$

$\mathsf{Consider,}$

$\mathsf{31+32+33+\;.\;.\;.\;.\;.+38}$

$\textsf{This can be written as,}$

$\mathsf{=1+2+3+\;.\;.\;.\;.+30+31+32+33+\;.\;.\;.\;.\;.+38-(1+2+3+\;.\;.\;.\;.+30)}$

$\mathsf{=(1+2+3+\;.\;.\;.\;.\;.\;.\;.+38)-(1+2+3+\;.\;.\;.\;.+30)}$

$\mathsf{=\dfrac{38(38+1)}{2}-\dfrac{30(30+1)}{2}}$

$\mathsf{=\dfrac{38{\times}39}{2}-\dfrac{30{\times}31}{2}}$

$\mathsf{=(19{\times}39)-(15{\times}31)}$

$\mathsf{=(19{\times}39)-(15{\times}31)}$

$\mathsf{=741-465}$

$\mathsf{=276}$

$\implies\boxed{\mathsf{31+32+33+\;.\;.\;.\;.\;.+38=278}}$

Answer 2

Given:

3 + 32 + 33 + ... + 38

To find:

The sum of the sequence of numbers 3 + 32 + 33 + ... + 38

Solution:

We must identify that,

• the later part of the sequence can be said to be an Arithmetic Progression (A.P)

3 + 32 + 33 + ... + 38

"Arithmetic Progression is a sequence where there is a consistent common difference between consecutive terms."

• In this section of the sequence, we can see that the common difference is 1.

32 (+1) , 33 (+1), 34 (+1), 35 (+1), 36 (+1), 37 (+1), 38 (+1)

• We now must find the sum of this section of sequence:

        To find the sum, use the formula for the sum of n terms

$S= \frac{n}{2} [2a + (n-1)d]$

d (common difference) = 1

n (number of terms in the sequence) = 7

a (first term) = 32

$S= \frac{7}{2} [2(32) + (7-1)1]\\S= 4.5 [64 + 6]\\S= 4.5 (70)\\S= 315$

• Now, add 3 to the sum to complete the sequence

3 + 32 + 33 + ... + 38 = 315 +3

                                  = 318

Hence,  3 + 32 + 33 + ... + 38 is 318.