Choose the correct answer from the alternatives given.If 12+22 +.... + x2 = 6x(x+1)(2x+1)then 12 + 32+ 52 + .... + 192 is equal to

(12+32+52+...........+192)= (1^2+2^2+3^2+...........+202) −(22+42+62+...........+202)= (12+22+32+..........+202)−4(1+22+32+..........+102)Using the formula,=620× 21× 41−4(610× 11× 21)=2870−1540=1330

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Choose the correct answer from the alternatives given. If 1^2+

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Choose the correct answer from the alternatives given.
If $1^{2} + 2^{2}$ +.... + $x^{2}$ = $\frac{x ( x + 1 ) ( 2 x + 1 )}{6}$ then $1^{2}$ + $3^{2}$ + $5^{2}$ + .... + $1 9^{2}$ is equal to

1330

2100

2485

2470

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Emily Discussed

Choose the correct answer from the alternatives given.
If

1^{2} + 2^{2}

+.... +

x^{2}

\frac{x ( x + 1 ) ( 2 x + 1 )}{6}

then

1^{2}

3^{2}

5^{2}

+ .... +

1 9^{2}

is equal to

14 mins ago

Discuss this question LIVE

14 mins ago

Text solutionVerified

(1^{2} + 3^{2} + 5^{2} + ........... + 1 9^{2})

(1^2

+

2^2

+

3^2

+ ........... + 2 0^{2}) - (2^{2} + 4^{2} + 6^{2} + ........... + 2 0^{2})

(1^{2} + 2^{2} + 3^{2} + .......... + 2 0^{2}) - 4 (1 + 2^{2} + 3^{2} + .......... + 1 0^{2})

Using the formula,

= \frac{20 \times 21 \times 41}{6} - 4 (\frac{10 \times 11 \times 21}{6}) = 2870 - 1540 = 1330

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Question Text	Choose the correct answer from the alternatives given. If $1^{2} + 2^{2}$ +.... + $x^{2}$ = $\frac{x ( x + 1 ) ( 2 x + 1 )}{6}$ then $1^{2}$ + $3^{2}$ + $5^{2}$ + .... + $1 9^{2}$ is equal to
Answer Type	Text solution:1
Upvotes	150

Choose the correct answer from the alternatives given.If 12+22 +.... + x2 = 6x(x+1)(2x+1)​then 12 + 32+ 52 + .... + 192 is equal to

Text solutionVerified

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Choose the correct answer from the alternatives given.
If $1^{2} + 2^{2}$ +.... + $x^{2}$ = $\frac{x ( x + 1 ) ( 2 x + 1 )}{6}$ then $1^{2}$ + $3^{2}$ + $5^{2}$ + .... + $1 9^{2}$ is equal to