The increasing sequence 1,4,5,16,17,20,21,64 include all powers of 4 and all the numbers that can be written as a sum of distinct powers of 4. Then the 63rd element of the sequence is:
Answers
The series can be written as:
40=1
41=4
41+40=4+1=5
42=16
42+41=16+4=20
42+41+40=16+4+1=21
So, in general, for 4n , there could be 4n,4n−1,4n−2,…,40=n+1 terms
So, total number of terms up to 4n = terms for 40 + terms for 41 + terms for 42+…+ terms for 4n
Sn=1+2+3+…+(n+1)=(n+1)(n+2)2
To find the 63rd term, let us find the n which covers 63 terms.
Sn=63=(n+1)(n+2)2
⟹(n+1)(n+2)=2∗63⟹n2+3n−124=0
⟹n=−3±32−4(1)(−124)−−−−−−−−−−−−−√2∗1
⟹n=−3+505−−−√2, ignoring the negative value for n
⟹n>−3+22.472>9
number of terms up to n = 9 =S9=(9+1)(9+2)2=55
t55=49+48+47+46+45+44+43+42+41+40
t56=410
t57=410+49
t58=410+49+48
t59=410+49+48+47
t60=410+49+48+47+46
t61=410+49+48+47+46+45
t62=410+49+48+47+46+45+44
t63=410+49+48+47+46+45+44+43=1398080
Ans: 1398080
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There is one correction needed in the question. The terms are 1,4,51,4,5, 16 and NOT 6.
The series can be written as:
40=140=1
41=441=4
41+40=4+1=541+40=4+1=5
42=1642=16
42+41=16+4=2042+41=16+4=20
42+41+40=16+4+1=2142+41+40=16+4+1=21
So, in general, for 4n4n, there could be 4n,4n−1,4n−2,…,40=n+14n,4n−1,4n−2,…,40=n+1 terms
So, total number of terms up to 4n4n = terms for 4040 + terms for 4141+ terms for 42+…+42+…+terms for 4n4n
Sn=1+2+3+…+(n+1)=(n
Step-by-step explanation:
So, in general, for 4n4n, there could be 4n,4n−1,4n−2,…,40=n+14n,4n−1,4n−2,…,40=n+1 terms
So, total number of terms up to 4n4n = terms for 4040 + terms for 4141+ terms for 42+…+42+…+terms for 4n4n
Sn=1+2+3+…+(n+1)=(n+1)(n+2)2Sn=1+2+3+…+(n+1)=(n+1)(n+2)2
To find the 63rd63rd term, let us find the n which covers 63 terms.
Sn=63=(n+1)(n+2)2Sn=63=(n+1)(n+2)2
⟹(n+1)(n+2)=2∗63⟹n2+3n−124=0⟹(n+1)(n+2)=2∗63⟹n2+3n−124=0
⟹n=−3±32−4(1)(−124)−−−−−−−−−−−