Latest TCS Aptitude Question SOLUTION: 1.2.3 + 2.3.4 + 3.4.5 + 4.5.6 + ..... + 127.128.129 find the sum

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(#M40024600) TCS Logical Reasoning question Keep an EYE

1.2.3 + 2.3.4 + 3.4.5 + 4.5.6 + ..... + 127.128.129 find the sum

Asked In TCS GAURAV JAISWAL (10 years ago)

Unsolved Read Solution (5)

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Read Solution (5) :

Aashmi chaudhary  (10 years ago)

the 1st term in the series is (1)(1+1)(1+2)
the 2nd term is (2)(2+1)(2+2)

therefore, the nth term is (n)(n+1)(n+2)
which when expanded gives (n^3 + 4n^2 + 4n)
here n is 127
Σ (n^3 + 3n^2 + 2n) [for n = 1 to n = 127]
Σ (n^3) + 3 Σ (n^2) + 2 Σ (n) [for n = 1 to n = 127]
after solving it we get 68153280
ans is 68153280

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ANIL AGRAHARI  (10 years ago)

this series can be written as sum(n.(n+1).(n+2))=sum(n^3)+sum(3n^2)+sum(2n)={n(n+1)/2}^2 +3{n(n+1)(2n+1)/6} +2{n(n+1)/2}= 68153280

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kalyani dangeti  (10 years ago)

@ rudra...
summation(n(n+1)(n+2))
=sum(n^3)+3sum(n^2)+2sum(n)
summation of n=n(n+1)/2 1st series
summation of n^2=n(n+1)(2n+1)/6 2nd series
summation of n^3=(n^2(n+1)^2)/4 3rd series n=127
hence ans=68153280

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Priyank Kumar  (10 years ago)

68153280...
we have 3 series 1,2,3
2,3,4'
3,4,5
find nth term of each individual series and multiply them to get nth term of series. use summation and put the value of n=127

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Rudra  (10 years ago)

@Priyank Kumar
for 1st series- sum of the series =127/2(1+127)
for 2nd series- sum of the series =127/2(2+128)
for 3rd series- sum of the series =127/2(3+129)
am i right???
if not plzzz correct me....

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