1,4,9,16,...... sum of this sequence upto 20 terms will be

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26 Sep 2011 sum

1,4,9,16,......
sum of this sequence upto 20 terms will be

OPtion

1) 1980
2) 2050
3) 2870
4) 2980
5) 3160
6) 3290
7) 3340
8) 3560
9) 3920
10) none of these

Solution

1+4+9+16+.....................
= 1^2+2^2+3^2+4^2+..............
= sum of squares of first 20 integers
according to sigma summation sum of squares of first n integers
= (1/6)*n*(n+1)*(2n+1)
on putting n = 20, answer will be obtained

Correct Option: 3 Best Solution (16)

Shiv   12 years AGO

sum = x(x+1)(2x+1)/6
put x=20

sum = 20*21*41/6= 2870

Like? Yes (9) | No   

Sukh   12 years AGO

1+4+9+16+.........+400 = 2870

Like? Yes (9) | No (2)  

E KESAVULU NAIDU   12 years AGO

add 1 to 2o squares i.e 1+4+9+.....20square(400)
so finally ans is:2870

Like? Yes (3) | No   

Rohit kumar   12 years AGO

its simple..... use the formula sum of square of numbers
i.e {n(n+1)(2n+1)}/6
put here the value of n=20,
then answer is 2870.

Like? Yes (3) | No   

A.P.Singh   12 years AGO

Sum of squares of first n numbers = n*(n+1)*(2n+1)/6
so Sum of squares of first 20 numbers = 20*21*41/6 = 2870

Like? Yes (3) | No (1)  

sowmya   12 years AGO

sum of squares of n terms=(n*(n+1)*(2n+1))/6
so sum of the given sequence=(20*21*41)/6
=2870.

Like? Yes (2) | No   

KIRAN B. MALI   12 years AGO

Sum of squares of first 'n' natural no's is,
= 1/6 n(n+1)(2n+1),
here n = 20;
therefore,
1/6 * 20(20+1)(40+1)= 1/6 * 17220
= 2870,
ans is 2870..

Like? Yes (2) | No   

ANUJ MAURYA   12 years AGO

the given series is square of numbers..
and we know that the sum of square of first n numbers= n(n+1)(2n+1)/6
= 20*21*41/6
= 2870

Like? Yes (2) | No   

Balakumaran N.A   12 years AGO

1^2+2^2+...n^2=n(n+1)(2n+1)/6
=20*21*41/6
=2870

Like? Yes (2) | No (1)  

Gunaseelan   12 years AGO

Series is 1^2 + 2^2 + 3^3+ .....+ 25 ^2
n=20
Sum of the series = n(n+1)(2n+1)/6
Result= (20 * 21 * 41)/6
= 2870

Like? Yes (2) | No (1)  

Ankit   12 years AGO

it is som of n square series

Like? Yes (1) | No   

SACHIN VERMA   12 years AGO

it a special series i.e.
1^2 2^2 3^2 ..............20^2
so sum of this series is given by
n(n+1)(2n+1)/6
where n is no of terms here n =20
sum= 20*21*41/6
= 2870

Like? Yes (1) | No   

kalirathinam   12 years AGO

(n*(n+1)*(2n+1))/6 sub 20in terms of n

Like? Yes (1) | No   

srilatha   12 years AGO

sum of squares of 1st 20terms will be
1+4+9+........+289+324+361+400=2870

Like? Yes (1) | No   

Uday Pratap Singh   12 years AGO

the solution is the series of squares of natural nos. upto 20 terms.

Like? Yes (1) | No   

YadeeshKumar   12 years AGO

1^2+2^2+...n^2=n(n+1)(2n+1)/6
=20*21*41/6
=2870

Like? Yes (1) | No (1)