Question

sum of 1,4,9,16 ...... 'n' terms is. A)n(n+1)(n+2)/6
B)n(n+1)(2n+1)/6
C)n(n+1)(2n+1)/12
D)n(n+1)(n+2)/12

Answer 1

Answer:

1,4,9,16,....

Explanation:

1 can be written as 1 square

4 can be written as 2 square

9 canbe written as 3 square

16 can be written as 4 square

summation n square=n(n+1)(2n+1)/6

as the above numbers are in squares we should use the above formula.

therefore option is B

Answer 2

Answer:

Sum of 1,4,9,16 ...'n' terms is n(n+1)(2n+1)/6.

Explanation:

• Sum of 1,4,9,16....n

That means

• The squares of natural numbers are: 1², 2², 3², 4²,…n².

• This be neither an AP nor GP since either the difference between two consecutive numbers is not constant or the ratio of two consecutive numbers is not constant. Consider the sum of this series by assuming an expression given below:

$k^{3}-(k-1)^{3}=3 k^{2}-3 k+1\\Substituting k=1 ,\\&1^{3}-(1-1)^{3}=3(1)^{2}-3(1)+1 \\&1^{3}-0^{3}=3(1)^{2}-3(1)+1 \ldots .(\mathrm{i})\end{aligned}$

$k^{3}-(k-1)^{3}=3 k^{2}-3 k+1Substituting k=1 ,&1^{3}-(1-1)^{3}=3(1)^{2}-3(1)+1 \\&1^{3}-0^{3}=3(1)^{2}-3(1)+1 \ldots .(\mathrm{i})\end{aligned}$

$Substituting k=2 ,&2^{3}-(2-1)^{3}=3(2)^{2}-3(2)+1 \\&2^{3}-1^{3}=3(2)^{2}-3(2)+1 \ldots . \text { (ii) }\end{aligned}$

$Substituting k=3 ,\\&3^{3}-(3-1)^{3}=3(3)^{2}-3(3)+1 \\&3^{3}-2^{3}=3(3)^{2}-3(3)+1 \ldots . \text { (iii) }\end{aligned}$

$Substituting k=4 ,\\&4^{3}-(4-1)^{3}=3(4)^{2}-3(4)+1 \\&4^{3}-3^{3}=3(4)^{2}-3(4)+1 \ldots \text { (iv) }\end{aligned}$

$Substituting k=n ,n^{3}-(n-1)^{3}=3(n)^{2}-3(n)+1$

Now, adding both sides of these equations together, we get;

$&1^{3}-0^{3}+2^{3}-1^{3}+3^{3}-2^{3}+\ldots+n^{3}-(n-1)^{3}=3\left(1^{2}+2^{2}+3^{2}+4^{2}+\ldots+n^{2}\right)-3(1+2+3+4+\ldots+n)+n(1)$

$&n^{3}-0^{3}=3\left(1^{2}+2^{2}+3^{2}+4^{2}+\ldots+n^{2}\right)-3(1+2+3+4+\ldots+n)+n\end{aligned}$

$n^{3}=3 \sum_{k=1}^{n} k^{2}-3 \sum_{k=1}^{n} k+n \\Here, \sum_{k=1}^{n} k$

represents the sum of first n natural numbers and is equal to

$\frac{n(n+1)}{2}\\So,n^{3}=3 \sum_{k=1}^{n} k^{2}-3\left[\frac{n(n+1)}{2}\right]+n$

Rearranging the terms,

$&\sum_{k=1}^{n} k^{2}=\frac{1}{3}\left[n^{3}+3\left[\frac{n(n+1)}{2}\right]-n\right] \\&\sum_{k=1}^{n} k^{2}=\frac{1}{6}\left[2 n^{3}+3 n^{2}+3 n-2 n\right] \\&=(1 / 6)\left(2 n^{3}+3 n^{2}+n\right) \\&=(1 / 6)\left[n\left(2 n^{2}+3 n+1\right)\right] \\&=(1 / 6)[n(n+1)(2 n+1)]$

Therefore, the sum of squares of first n natural numbers $=\frac{n(n+1)(2 n+1)}{6}$