Question

Find the general term of the following sequence:
1, 5, 15, 34, 65, ...

Hint:
(1), (2+3), (4+5+6), (7+8+9+10), ...​

Answer 1

Step-by-step explanation:

1 = 1

2+3 = 5

4+5+6 = 15

7+8+9+10 = 34

11+12+13+14+15 = 65

16+17+18+19+20+21 = 111

22+23+24+25+26+27+28 = 175

THIS IS THE GENERAL FORM OF THE FOLLOWING SEQUENCE BRO.

HOPE IT HELPS YOU.

Answer 2

Here we have to find the general term of the sequence given by :-

• 1 , 5 , 34 , 65 , . . . .

Can be further written as ;

• 1 , (2+3) , (4+5+6) , (7+8+9+10) , . . .

Really , a very very complex sequence . Neither Arithmetic nor Geometric nor harmonic !!!!

At first see the sequence you will find that , any term in the sequence is the sum of natural numbers as same as the term ( numbers not in order ) . Let's consider 3rd term , it is sum of 3 natural number not taken in order and so on all terms . . . . . .

Now , very very carefully see the attachment no. 1 and 2, you will get your answer finally :)

Answer is :-

• ${ \pmb { \blue { \bf { T_{n} = \displaystyle \sum \bf x - \sum ( \bf x - n ) }}}}$

Where , ${ \pmb { \green { \bf { x = \displaystyle \sum n }}}}$ and ${ \pmb { \green { \bf { \displaystyle \sum n = \dfrac{n(n+1)}{2} }}}}$

Note :- I got many general terms for this series , but they were inapplicable on 1st term . But this formula also works on 1st term :)

When you put n = 1 , for 1st term then you get ;

${ \pmb { \green { \bf { \displaystyle \sum 0 = \dfrac{0(0+1)}{2} = \dfrac{0 (1)}{2}= \dfrac{0}{2} = 0 }}}}$

Then you get , 1st term = 1 - 0 = 1 . which also satisfies the formula :)

And sorry for the handwriting :)"