Latest TCS Aptitude Question SOLUTION: (11^2 + 12^2 + 13^2 + ... + 20^2) = ?

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(#M40019256) TCS question Keep an EYE

(11^2 + 12^2 + 13^2 + ... + 20^2) = ?

Asked In TCS Kalidindi Mounika (10 years ago)

Unsolved Read Solution (7)

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Aditya Solge  (10 years ago)

The sum of squares of first n natural numbers is:

(n*(n+1)*(2n+1))/6

Thus find out the sum of squares of first 20 natural numbers(1-20) i.e 2870
and now the sum squares of first 10 natural numbers(1-10) is : 385

and the difference between these two will give you the sum of (11^2 + 12^2 + 13^2 + ... + 20^2) which is equal to 2485

Is this solution Helpfull? Yes (45) | No   

Ish Mishra  (10 years ago)

Okay So here we go:

Re-write this as
(10+1)^2+(10+2)^2+(10+3)^2...+(10+10)^2
This would give us:

10*10^2 + (1^2+2^2+3^2+..+10^2) + (20+40+60+..200)

So now you get:
1000+Sum of square 10 natural no.+ Sum of an A.P for 10 terms

Answer:2485

Is this solution Helpfull? Yes (12) | No (4)  

puja  (10 years ago)

(11^2 + 12^2 + 13^2 + ... + 20^2) = (1^2+2^2+.............20^2)-(1^2+2^2+.........10^2)
{(20*21*41)/6-(10*11*12)/6}=(2870-385)=2485

Is this solution Helpfull? Yes (6) | No (2)  

pria rajan  (10 years ago)

first of all we should be clear of square numbers..
11^2=121 12^2=144 13^2=169 14^2=196 15^2=225 16^2=256 17^2=289 18^2=324 19^2=361 20^=400
121+144+169+196+225+256+289+324+361+400=2485
this is easy to find square of numbers: for eg: xy^2= you must first square the ending digit y and if it comes as double digit number then we must add that to+ xy+y
eg: 14^2
4*4=16
hence keep that 6 as it is
now 14+4+1=19 hence the answer is 196

Is this solution Helpfull? Yes (2) | No (4)  

Saloni Singh  (10 years ago)

121+144+169+196+225+256+289+324+361{as i hav learnt da squares}
=2085{added horizontally rather dan vertically writing inorder to save time}

Is this solution Helpfull? Yes | No (15)  

richa ahuja  (10 years ago)

(10+1)^2+(10+2)^2+(10+3)^2+....+(10+10)^2
10^2+1^2+2*10*1+10^2+2^2+2*10*2+....+10^2+10^2+2*10*10
(10^2*10)+(1^2+2^2+3^2+...10^2)+2*10*(1+2+3+...+10)
sum of n natural no=(n(n+1))/2
sum of squares of n natural nos=n(n+1)(2n+1)/6
here n=10
1000+385+55=1440

Is this solution Helpfull? Yes | No (2)  

Manisha Gupta  (10 years ago)

value=2485

Is this solution Helpfull? Yes | No   

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