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[22+ 32 + 42 + 52 + 62 + 72 + 82 + 92 + 102 + 112] is equal to
- A506
- B2916
- C540
- D505
Answer not set
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0
CCommunity
The given expression:
[22+ 32 + 42 + 52 + 62 + 72 + 82 + 92 + 102 + 112]
= [12 +22+ 32 + 42 + 52 + 62 + 72 + 82 + 92 + 102 + 112] – 1
= The sum of squares of first 11 natural numbers – 1
[We know that, the sum of square(s) of 1st n natural numbers = $(\frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6})$
Here n = 11]
= The sum of squares of first 11 natural numbers – 1
$(= \frac{{11\left( {11 + 1} \right)\left( {22 + 1} \right)}}{6} - 1)$
= 506 – 1
= 505
[22+ 32 + 42 + 52 + 62 + 72 + 82 + 92 + 102 + 112]
= [12 +22+ 32 + 42 + 52 + 62 + 72 + 82 + 92 + 102 + 112] – 1
= The sum of squares of first 11 natural numbers – 1
[We know that, the sum of square(s) of 1st n natural numbers = $(\frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6})$
Here n = 11]
= The sum of squares of first 11 natural numbers – 1
$(= \frac{{11\left( {11 + 1} \right)\left( {22 + 1} \right)}}{6} - 1)$
= 506 – 1
= 505
0
CCommunity
The given expression:
[22+ 32 + 42 + 52 + 62 + 72 + 82 + 92 + 102 + 112]
= [12 +22+ 32 + 42 + 52 + 62 + 72 + 82 + 92 + 102 + 112] – 1
= The sum of squares of first 11 natural numbers – 1
[We know that, the sum of square(s) of 1st n natural numbers = $(\frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6})$
Here n = 11]
= The sum of squares of first 11 natural numbers – 1
$(= \frac{{11\left( {11 + 1} \right)\left( {22 + 1} \right)}}{6} - 1)$
= 506 – 1
= 505
[22+ 32 + 42 + 52 + 62 + 72 + 82 + 92 + 102 + 112]
= [12 +22+ 32 + 42 + 52 + 62 + 72 + 82 + 92 + 102 + 112] – 1
= The sum of squares of first 11 natural numbers – 1
[We know that, the sum of square(s) of 1st n natural numbers = $(\frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6})$
Here n = 11]
= The sum of squares of first 11 natural numbers – 1
$(= \frac{{11\left( {11 + 1} \right)\left( {22 + 1} \right)}}{6} - 1)$
= 506 – 1
= 505