Question

Find the sum to n terms of the series
${3 }^{2} + {6}^{2} + {9}^{2} .......... {30}^{2}$
Answer-3465
Solution-?​

Answer 1

$\huge\tt\blue{Solution}$

Here,

a=

${3}^{2} \\ = 9$

d= ${6}^{2} - {3}^{2}$

$= 36 - 9$

$= 27$

So,

$nth \: term \: an = {30}^{2} \\ = > a + (n - 1) \times d = 900 \\ = > 9 + (n - 1) \times 27 = 900 \\ = > (n - 1) \times 27 = 900 - 9 \\ = > n - 1 = \frac{891}{27} \\ = > n = 33 + 1 \\ = > n = 34$

Again,

sum upto nth term, Sn

$= \frac{n}{2} (2a + (n - 1)d) \\ = \frac{34}{2} (2 \times 9 + 33 \times 27) \\ = 17(18 + 891) \\ = 17 \times 909 \\ = 15453$

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Answer 2

Answer:

S=1⋅2⋅3+2⋅4⋅6+3⋅6⋅9+⋯

=∑n=1Nn(2n)(3n)=∑n=1N6n3=6

[N(N+1)/2]2=3N2(N+1)2/2