(C) We need to evaluate the sum $\sum_{n=1}^5 (n^3 + n^2 + n)$.For $n=1$: $1(1^2+1+1) = 1(3) = 3$.For $n=2$: $2(2^2+2+1) = 2(7) = 14$.For $n=3$: $3(3^

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(C) We need to evaluate the sum $\sum_{n=1}^5 (n^3 + n^2 + n)$.For $n=1$: $1(1^2+1+1) = 1(3) = 3$.For $n=2$: $2(2^2+2+1) = 2(7) = 14$.For $n=3$: $3(3^2+3+1) = 3(13) = 39$.For $n=4$: $4(4^2+4+1) = 4(21) = 84$.For $n=5$: $5(5^2+5+1) = 5(31) = 155$.Summing these values: $3 + 14 + 39 + 84 + 155 = 295$.

Class 11 Mathematics Sequences and Series nth term of special series, Sum to n terms and Infinite number of terms

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$\sum_{n=1}^5 n(n^2+n+1) = $

AP EAMCET 2018 All AP EAMCET

A
$500$
B
$155$
C
$295$
D
$395$

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Sequences and Series

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nth term of special series, Sum to n terms and Infinite number of terms

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AP EAMCET 2018 Paper All AP EAMCET years →

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$\sum_{n=1}^5 n(n^2+n+1) = $

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