the value of (1^2 + 2^2 + 3^2 + ...+n^2) will be: the options are below:?

Question

1) n(n+1)/2
2)[n(n+1)(2n+1)]/6
3)[n(n+1)(2n-1)]/6
4)[n(n+1)/2)]^2

the right option is the 2nd one, but i dunno why.. please show ur working out:) thanks

Answer 1

we need to prove this by induction...

for n belonging to Z

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

for n=1

LHS:
f(1)= 1^2 =1
RHS
1*(1+1)8(2*1+1)/6=6/6=1

Thus its true when n=1;

Assume f(k) is true i.e.
1^2+2^2+......+k^2=k(k+1)(2k+1).. Let this be equation 1

Now we need to prove f(k+1) is true...
ie
f(k+1)=(k+1)(k+2)(2(k+1)+1)/6
=(k+1)(k+2)(2k+3)/6
L.h.s

f(k+1)=1^2+2^2........+k^2+(k+1)^2
=k(k+1)(2K+1)/6 + (k+1)^2 from eq 1
=(k+1)(k(2k+1)/6 +k+1)
=(1/6)(k+1)(2k^2+k+6k+6)
=(1/6)(k+1)(2k^2+7k+6)
=(1/6)(k+1)(k+2)(2k+3)

Hence f(k+1) is true if f(k) is true....

Thus its true for all positive integers.......

Answer 2

There are normally 2 questions regarding the sum of squares series: the proving and the derivative of the series. I'll cover the derivative, 'cause I think that's your question.

First by working out the first few "sums" in the series, we get:
n = 1: Sum = 1^2 = 1
n = 2: Sum = 1^2 + 2^2 = 5
n = 3: Sum = 1^2 + 2^2 + 3^2 = 14
:
:
And these sums form a number pattern as follow:
1, 5, 14, 30, 55, 91, ...
4, 9, 16, 25, 36, .... 1st difference
5, 7, 9, 11, .... 2nd difference
2, 2, 2, .... 3rd difference

If we are able to find the general formula for the nth term of this number pattern, we will actually obtain the general formula for the sums of squares.

Using the method of finite difference (ie, taking the difference of any consecutive terms in a pattern), we can observe from the above that the number pattern comes to a common difference at the 3rd row, meaning that the formula we're looking is an order-3 or cubic equation, so we first assume the equation to be the gerealised form:
an^3 + bn^2 + cn + d, where a, b, c, d are constants we need to find. This can be done by constructing 4 simultaneous equations using the first 4 numbers of the series, I'll show you 2 of them, hopefully you can get the other 2.

When n = 1 (ie, first term), a(1)^3 + b(1)^2 + c(1) + d = 1
=> a + b + c + d = 1 ---------------(1)
When n = 2, a(2)^3 + b(2)^2 + c(2) + d = 5
=> 8a + 4b + 2c + d = 5 ----------(2)
:
:

Solving gives a=1/3, b=1/2, c=1/6, and d=0
Thus, the formula is 1/3n^3 + 1/2n^2 + 1/6n, factorising yields
=>1/6 (n) (2n^2 + 3n + 1)
=>1/6 (n) (n + 1) (2n + 1)

For the proving part, you can look up the method of Mathematical Induction. If you need help on this also, drop a line.

Answer 3

Use mathematical induction to prove that

1² + 2² + 3² + ... + n² = n(n + 1)(2n + 1)/6



___________

3 steps

___________

STEP1: Prove that it is true for n = 1



If n = 1, then we can easily see that it is true:

1² = 1(1 + 1)[2(1) + 1]/6

1 = 1(2)(3)/6

1 = 1



_____________

STEP2: Assuming n = k is true, then we have to prove that n = k + 1 is also true.



Now, assuming n = k is true,

1² + 2² + 3² + ... + k² = k(k + 1)(2k + 1)/6



If we add (k + 1)² to both sides,

1² + 2² + 3² + ... + k² + (k + 1)² = k(k + 1)(2k + 1)/6 + (k + 1)²



Now, we can factor out k + 1 from the expression at the right

1² + 2² + 3² + ... + (k + 1)² = (k + 1)[k(2k + 1)/6 + k + 1]



We simplify

1² + 2² + 3² + ... + (k + 1)² = (k + 1)(2k² + k + 6k + 6)/6



Or

1² + 2² + 3² + ... + (k + 1)² = (k + 1)(2k² + 7k + 6)/6



We factor out the right side

1² + 2² + 3² + ... + (k + 1)² = (k + 1)(k + 2)(2k + 3)/6



Then,

1² + 2² + 3² + ... + (k + 1)² = (k + 1)[(k + 1) + 1][2(k + 1) + 1]/6



We get the same statement as assuming that n = k + 1 is true.



__________

STEP3:

Using the principles of mathematical induction, we therefore conclude that the equation is true for all values of n.

^_^

Answer 4

The answer is option:1) n(n+1)/2