Algebra> Find the sum of n terms 1 square +( 1 squ...

Find the sum of n terms 1 square +( 1 square2 suare) +( 1square +2 square + 3 square) +

Deepika k gowda , 5 Years ago

Grade 11

2 Answers

Vikas TU

Last Activity: 5 Years ago

Dear student

Please folow the below link to get your ans

https://www.askiitians.com/forums/Algebra/1square-1square-2square-1square-2square-3square_192192.htm

Good Luck

Onkar mamidwar

Last Activity: 2 Years ago

(12)+(12+22)+...+(12+22+...+n2)=n(n+1)2(n+2)12

Explanation:

We will use the following known sums (each of which can be proven via induction):

n∑i=1i=n(n+1)2
n∑i=1i2=n(n+1)(2n+1)6
n∑i=1(i3)=n2(n+1)24

With those:

(12)+(12+22)+...+(12+22+...+n2)

=n∑i=1(12+22+...+i2)
=n∑i=1i∑j=1j2
=n∑i=1i(i+1)(2i+1)6
=n∑i=12i3+3i2+i6
=n∑i=113i3+n∑j=112j2+n∑k=116k
=13n∑i=1i3+12n∑j=1j2+16n∑k=1k
=13(n2(n+1)24)+12(n(n+1)(2n+1)6)+16(n(n+1)2)
=n2(n+1)212+n(n+1)(2n+1)12+n(n+1)12
=n(n+1)[n(n+1)+(2n+1)+1]12
=n(n+1)(n2+3n+2)12
=n(n+1)2(n+2)12
(12)+(12+22)+...+(12+22+...+n2)=n(n+1)2(n+2)12
Explanation:
We will use the following known sums (each of which can be proven via induction):

n∑i=1i=n(n+1)2

n∑i=1i2=n(n+1)(2n+1)6

n∑i=1(i3)=n2(n+1)24
With those:
(12)+(12+22)+...+(12+22+...+n2)
=n∑i=1(12+22+...+i2)
=n∑i=1i∑j=1j2
=n∑i=1i(i+1)(2i+1)6
=n∑i=12i3+3i2+i6
=n∑i=113i3+n∑j=112j2+n∑k=116k
=13n∑i=1i3+12n∑j=1j2+16n∑k=1k
=13(n2(n+1)24)+12(n(n+1)(2n+1)6)+16(n(n+1)2)
=n2(n+1)212+n(n+1)(2n+1)12+n(n+1)12
=n(n+1)[n(n+1)+(2n+1)+1]12
=n(n+1)(n2+3n+2)12
=n(n+1)2(n+2)12