Question

the sequence 1,2,1,2,2,1,2,2,2,1,2,2,2,2,1,2,2,2,2,2,1...........so on. find the sum of 1234 terms of this sequence ?​

Answer 1

Answer:

As we have to calculate the sum of this series , we will write the series whose sum would be same as this series. Here we will represent the whole no. Of 2′s between 1 as a single term which is sum of all those 2′s present between two 1′s. Series would be as follows

1,2,1,4,1,6,1,8,1....... m times

Now to evaluate the no. Of 1′s , we 1st calculate value of m. Now we write another series to evaluate value of m, we write a single term between all 1′s , but this term would be equal to the no. Terms between the 1′s including the latter 1

For 2,2,1 =3 terms ; for 2,2,2,1=4 terms and so on

We have a new series where we have no. of terms

1,2,3,4,5,6...... n times

Here this series is in AP and sum of this is series 1230 ( total no. Of terms )

Also in each term we have one 1 term present , so total no. Of 1′s present is equal to n

Now

1230= n/2*(2*1+(n-1)*1) , now we get the solution n= 49.1 which is shows that last terms is not one and sum of this series upto n=49 is 1225 therefore there are more 5 terms equal to 2 which have not terminated by 1 . There fore there are total 49 no. Of 1′s

Now coming back to our equation we now have

1,2 ,1,4,1,6......97times 2,2,2,2,2

as we know last 5 terms are not terminated by 1 therefore now we have

49*1 + 48/2*(2*2+(48–1)*2) + 10

Where 49*1 is sum of all 1′s which are 49 in no.

Second term in above equation is sum of AP of common difference 2 and

Last term 10 is sum of last 5 terms which are 2′s

Sum of this series comes out to be 2411

Hence the solution