The number of values of n- for which pn-1- -2- -3- -4- - n- is the square of a natural number- is equal to

The sum of squares of first n natural numbers is given by $\frac{1}{6} n (n + 1) (2 n + 1)$ or $\frac{1}{6} (2 n^{3} + 3 n^{2} + n)$ . Find the sum of squares of the first 10 natural numbers.

Q. Use the principle of mathermatical induction to prove that

1^{3} + 2^{3} + . . . . . n^{3} = \frac{1}{4} n^{2} {(n + 1)}^{2}

for every natural number

n

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The number of values of n, for which p(n)=1!+2!+3!+4!+⋯+n! is the square of a natural number, is equal to

The correct option is C 2For n=4, P(n)=1+2+6+24=33.For n>4, n! is always having 0 in the units place.So the n=1,3 are the answers.

The number of values of $n$ , for which $p (n) = 1! + 2! + 3! + 4! + \dots + n!$ is the square of a natural number, is equal to

The correct option is C $2$
For $n = 4$ , $P (n) = 1 + 2 + 6 + 24 = 33$ .
For $n > 4$ , $n!$ is always having $0$ in the units place.
So the $n = 1, 3$ are the answers.