If 12-a1-22-a2-32-a3-n2-an-1-3 nn2-1- then an is given by

The correct option is A n (1^2 a_1)+(2^2 a_2)+(3^2 a_3)+.......(n^2 a_n)=1/3 n(n^2 1) #160; (1^ 2 +2^ 2 +3^ 2 +....+n^ 2 ) (a_ 1 +a_ 2 +...+a_ ...

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Logarithmic Function

If 12-a1+22...

Question

If $(1^{2} - a_{1}) + (2^{2} - a_{2}) + (3^{2} - a_{3}) + . . . . . . . (n^{2} - a_{n}) = \frac{1}{3} n (n^{2} - 1)$ , then $a_{n}$ is given by

n

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n^{2}

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2 n

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\frac{n}{2}

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Solution

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a_{1}, a_{2}, a_{3}, . . . . . . . . . ., a_{n}

A . P .

(1^{2} - a_{1}) + (2^{2} - a_{2}) + (3^{2} - a_{3}) + . . . . . . . + (n^{2} - a_{n}) = \frac{(n - 1) n (n + 1)}{3}

(\frac{a_{5} + a_{3} - a_{2}}{6})

(1^{2} - a_{1}) + (2^{2} - a_{2}) + (3^{2} - a_{3}) + \dots (n^{2} - a_{n}) = \frac{1}{3} n (n^{2} - 1),

a_{7} .

(1^{2} - a_{1}) + (2^{2} - a_{2}) + . . . + (n^{2} - a_{n}) = \frac{1}{3} n (n^{2} - 1)

a_{n} = n + 1

1^{2} + 2^{2} + . . . + n^{2} = \frac{n (n + 1) (2 n + 1)}{6}

If $S_{n} = \frac{1}{1^{2}} + \frac{1}{2^{2}} + \frac{1}{3^{2}} + . . . . . + \frac{1}{n^{2}}$ and $T_{n} = \frac{2 n - 1}{n}$ , then

n^{t h}

a_{n}

a_{n} = a_{n - 1} + 2 n

n \geq 2

a_{1} = 1

a_{n}

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